NCERT Exemplar Q & A

Hint

(d) Key concept: A capacitor offers zero resistance in a circuit when it is uncharged, i.e., it can be assumed as short circuited and it offers infinite resistance when it is fully charged.

At steady state the capacitor offers infinite resistance in DC circuit and acts as open circuit as shown in figure, therefore no current flows through the capacitor and \(10 \Omega\) resistance, leaving zero potential difference across \(10 \Omega\) resistance. Hence potential difference across capacitor will be the potential difference across A and B .
The potential difference across lower and middle branch of circuit is equal to the potential difference across capacitor of upper branch of circuit.
Current flows through \(2 \Omega\) resistance from left to right, is given by \(I=\mathrm{V} / (\mathrm{R}+\mathrm{r})=1 \mathrm{~A}\). The potential difference across \(2 \Omega\) resistance, \(\mathrm{V}=\mathrm{IR}=1 \times 2=2 \mathrm{~V}\) Hence potential difference across capacitor is also 2 V. The charge on capacitor is \(\mathrm{q}=\mathrm{CV}=(2 \mu \mathrm{~F}) \times 2 \mathrm{~V}=8 \mu \mathrm{C}\).

Hint

(c) Key concept: Electric potential decreases in the direction of electric field. The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.
The positively charged particle experiences electrostatic force along the direction of electric field, hence moves in the direction of electric field. Thus, positive work is done by the electric field on the charge. We know
\(
W_{\text {electrical }}=-\Delta U=-q \Delta V=q\left(V_{\text {initial }}-V_{\text {final }}\right)
\)
Hence electrostatic potential energy of the positive charge decreases.

Hint

(c) As the direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential than other equipotential surface maintained at low electrostatic potential. Hence direction of electric field is from \(B\) to \(A\) in all three cases.
The positively charged particle experiences electrostatic force along the direction of electric field, hence moves in the direction opposite to electric field. Thus, the work done by the electric field on the charge will be negative. We know
\(
W_{\text {electrical }}=-\Delta U=-q \Delta V=q\left(V_{\text {initial }}-V_{\text {final }}\right)
\)
Here initial and final potentials are same in all three cases and same charge is moved, so work done is same in all three cases.

Hint

(c) We know, the electric field intensity E and electric potential V are related
\(\mathrm{E}=\mathrm{dV} / \mathrm{dr}\)
If electric field intensity \(\mathrm{E}=0\), then \(\mathrm{dV} / \mathrm{dr}=0\). It means, \(\mathrm{E}=0\) inside the charged conducting sphere causes uniform potential inside the sphere. Hence uniform electrostatic potential 100 V will be at any point inside the sphere.
Important points:

The electric field zero does not necessary imply that electric potential is zero. E.g., the electric field intensity at any point inside the charged spherical shell is zero but there may exist non-zero electric potential.
If two charged particles of same magnitude but opposite sign are placed, the electric potential at the midpoint will be zero but electric field is not equal to zero.

Hint

(a) The collection of charges, whose total sum is not zero, with regard to great distance can be considered as a single point charge. The equipotential surfaces due to a point charge are spherical.

Important point:
The electric potential due to point charge \(q\) is given by \(V=q / 4 \pi \epsilon_0 r\)
It means electric potential due to point charge is same for all equidistant points. The locus of these equidistant points, which are at same potential, form spherical surface.

Hint

(c) Here the system can be considered as two capacitors \(\mathrm{C}_1\) and \(\mathrm{C}_2\) connected in series as shown in figure.

The capacitance of parallel plate capacitor filled with dielectric block has thickness \(d_1\) and dielectric constant \(\mathrm{K}_2\) is given by
\(
\begin{aligned}
& \frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_1}+\frac{1}{C_2} \Rightarrow C_{\mathrm{eq}}=\frac{C_1 C_2}{C_1+C_2} \\
& C_{\mathrm{eq}}=\frac{\frac{K_1 \varepsilon_0 A}{d_1}}{\frac{K_1 \varepsilon_0 A}{d_1}+\frac{K_2 \varepsilon_0 A}{d_2}} \frac{K_2 \varepsilon_0 A}{d_2}=\frac{K_1 K_2 \varepsilon_0 A}{K_1 d_2+K_2 d_1} \dots(i)
\end{aligned}
\)
We can write the equivalent capacitance as
\(
C=\frac{K \varepsilon_0 A}{d_1+d_2} \dots(ii)
\)
On comparing (i) and (ii) we have
\(
K=\frac{K_1 K_2\left(d_1+d_2\right)}{K_1 d_2+K_2 d_1}
\)

Hint

(b c, d) We know, the electric field intensity E and electric potential V are
\(
E=-\frac{d V}{d r}
\)

Electric potential decreases inf the direction of electric field. The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.
The electric field in z-direction suggest that equipotential surfaces are in \(x-y\) plane. Therefore the potential is a constant for any x for a given z , for any y for a given z and on the \(x-y\) plane for a given z .

Hint

(a, b, c) The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density, larger the field strength.
We know, the electric field intensity E and electric potential V are related as a, b and c
We know that on any two points of equipotential surface, potential difference is zero or of equal potential.
\(
\because E=\frac{-d V}{d r}
\)
So the electric field intensity is inversely proportional to the separation between equipotential surfaces.
So equipotential surfaces are closer in regions of large electric. Thus, it verifies answer a
The electric field is larger near the sharp edge, due to larger charge density as a is very small
\(
\because \sigma=\frac{q}{A}
\)
So equipotential surfaces are closer or crowded. It verifies answer \(b\).
As the electric field \(E=\frac{k q}{r^2}\) and potential or field decreases as size of the body increases or vice-versa (case of the earth), so the equipotential surfaces will be more crowded if the charge density \(\sigma=\frac{q}{A}\) increases. It verifies the answer c .
As the equipotential surface depends on distance r by \(E=\frac{-d V}{r}\) and \(V=\frac{k q}{r}\). Equipotential surface depends on charge density at that place which is different at a different place, so equipotential surface are not equispaced all over.
Hence the electric field intensity E is inversely proportional to the separation between equipotential surfaces. So, equipotential surfaces are closer in regions of large electric fields. As electric field intensities is large near sharp edges of charged conductor and near regions of large charge densities. Therefore, equipotential surfaces are closer at such places.

Hint

(b, c) Key concept: The work done by the external agent in shifting the test charge along the dashed line from 1 to 2 is

The external agent does a work \(W=-q \int_1^2 \vec{E} \cdot d \vec{l}\) in transporting the test charge \(q\) slowly from the positions 1 to 2 in the static electric field.
\(
W_{\mathrm{ext}}=\int_1^2 \vec{F}_{\mathrm{ext}} \cdot \overrightarrow{d l}=\int_1^2(-q \vec{E}) \cdot \overrightarrow{d l}=-q \int_1^2 \vec{E} \cdot \overrightarrow{d l}
\)
We know \(V_A-V_B=-\int_A^B \vec{E} \cdot \overrightarrow{d l} \dots(i)\)
\(W_{\text {electrical }}=-\Delta U=-q \Delta V=q\left(V_A-V_B\right) \dots(ii)\)
Hence from (i) and (ii), \(W_{\text {electrical }}=q\left(V_A-V_B\right)=-q \int_A^B \vec{E} \cdot \overrightarrow{d l}\)
If we want to calculate the work done to move a charge along an equipotential from \(A\) to \(B\),
For equipotential surface \(V_{A}=V_B\), hence \(W=0\).
Also electric field is perpendicular to equipotential surface, hence
\(
\vec{E} \cdot \overrightarrow{d l} \Rightarrow W_{\text {electrical }}=0
\)

Hint

(b, c) We know, the electric field intensity E and electric potential V are dV
related as \(E=-d V / d r\)
or we can write \(|E|=\Delta V / \Delta r\)
The electric field intensity E and electric potential V are related as \(\mathrm{E}=0\) and for \(\mathrm{V}=\mathrm{constant} \mathrm{dV} / \mathrm{dr}=0,\), this imply that electric field intensity \(\mathrm{E}=0\).
If some charge is present inside the region then electric field cannot be zero at that region, for this \(\mathrm{V}=\) constant is not valid.

Hint

(a, d) Initially key \(K_1\) is closed and key \(K_2\) is open, the capacitor \(\mathrm{C}_1\) is charged by battery and capacitor \(\mathrm{C}_2\) is still uncharged. Now \(\mathrm{K}_1\) is opened and \(\mathrm{K}_2\) is closed, the capacitors \(\mathrm{C}_1\) and \(\mathrm{C}_2\) both are connected in parallel. The charge stored by capacitor \(\mathrm{C}_1\), gets redistributed between \(\mathrm{C}_1\) and \(\mathrm{C}_2\) till their potentials become same, i.e., \(\mathrm{V}_2=\mathrm{V}_1\).
By law of conservation of charge, the charge stored in capacitor Cx is equal to sum of charges on capacitors \(\mathrm{C}_1\) and \(\mathrm{C}_2\) when \(\mathrm{K}_1\) is opened and \(\mathrm{K}_2\) is closed, i.e.,
\(
Q_1^{\prime}+Q_2^{\prime}=Q
\)

Hint

(a, b) The potential of a body is due to charge of the body and due to the charge of surrounding. If tfiere are no charges anywhere else outside, then the potential of the body will be due to its own charge. If there is a cavity inside a conducting body, then charge can be placed inside the body. Hence there must be charges on its surface or inside itself. Hence option (a) is correct. The charge resides on the outer surface of a closed charged conductor. Hence there cannot be any charge in the body of the conductor. Hence option (b) is correct.

Hint

(c, d) The battery maintains the potential difference across connected capacitor in every circumstance. However, charge stored by disconnected charged capacitor remains conserved.
Case A: When key K is kept closed and plates of capacitors are moved apart using insulating handle. The battery maintains the potential difference across connected capacitor in every circumstance. The separation between two plates increases which in turn decreases its capacitance ( \(C=\epsilon_0 A / d\) )and potential difference across
connected capacitor continue to be the same as capacitor is still connected with battery. Hence, the charge stored decreases as \(\mathrm{Q}=\mathrm{CV}\).
Case B: When key K is opened and plates of capacitors are moved apart using insulating handle. The charge stored by isolated charged capacitor remains conserved. The separation between two plates is increasing which in turn decreases its capacitance with the decrease of capacitance, potential difference V increases as \(V=Q / C\).

Hint

(d) The capacitor has charges \(+Q\) and \(-Q\) on its plates.
The total charge enclosed is the sum of these charges:
\(Q_{\text {enc }}=+Q+(-Q)=0\)
The electric flux \(\Phi_E\) through the closed surface is given by:
\(\Phi_E=\frac{Q_{\text {enc }}}{\varepsilon_0}\)
Substitute \(Q_{\text {enc }}=0\) :
\(\Phi_E=\frac{0}{\varepsilon_0}=0\)
The electric flux through the closed surface is zero.

Hint

(d) The formula for equivalent capacitance in series is:
\(\frac{1}{C_{e q}}=\frac{1}{C_1}+\frac{1}{C_2}\)
Since \(C_1=C_2=C\) :
\(\frac{1}{C_{e q}}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C}\)
Therefore:
\(C_{e q}=\frac{C}{2}\)
The breakdown voltage for capacitors in series is the sum of individual breakdown voltages:
\(V_{e q}=V_1+V_2\)
Since \(V_1=V_2=V\) :
\(V_{e q}=V+V=2 V\)
The equivalent capacitance is \(\frac{C}{2}\) and the breakdown voltage is \(2 V\).

Hint

The equivalent capacitance of capacitors in parallel is the sum of their individual capacitances.
\(
\begin{aligned}
& C_{e q}=C+C \\
& C_{e q}=2 C
\end{aligned}
\)
In a parallel connection, the voltage across each capacitor is the same and is equal to the applied voltage \(V\). Therefore, the breakdown voltage of the combination remains the same as that of each capacitor:
\(
V_{\text {breakdown }}=V
\)
The equivalent capacitance is \(2 C\) and the breakdown voltage is \(V\).

Hint

(b) Two capacitors are in parallel combination.
Hence equivalent capacitance \(C_{\text {eq }}=C_1+C_2=C+C=2 C\)

Hint

(c) Step 1: An isolated capacitor has a fixed charge \((Q)\) on its plates. The force between the plates of a capacitor can be expressed using the formula:
\(
F=\frac{Q^2}{2 A \epsilon_0}
\)
where:
\(F\) is the force between the plates,
\(Q\) is the charge on the plates,
\(A\) is the area of the plates,
\(\epsilon_0\) is the permittivity of free space.
Step 2: Insert the Dielectric Slab
When a dielectric slab is inserted between the plates of the capacitor, it affects the electric field between the plates. However, since the capacitor is isolated, the charge \((Q)\) on the plates remains constant.
Step 3: Analyze the Effect of the Dielectric
The presence of the dielectric slab increases the capacitance of the capacitor. The new capacitance \(C^{\prime}\) can be expressed as:
\(
C^{\prime}=K \cdot C
\)
where \(K\) is the dielectric constant of the material and \(C\) is the original capacitance without the dielectric.
Step 4: Determine the New Force
Even though the capacitance increases, the charge remains constant. The force between the plates can still be expressed using the original formula:
\(
F=\frac{Q^2}{2 A \epsilon_0}
\)
Since \(Q[latex] and [latex]A\) are constant and \(\epsilon_0\) does not change, the force remains unchanged.
Thus, the force between the plates of an isolated capacitor with a dielectric slab inserted remains the same.

Hint

(d) The electric field due to a point charge is given by \(E=\frac{k Q}{r^2}\).
The energy density of an electric field is given by \(u=\frac{1}{2} \epsilon_0 E^2\).
The energy density is:
\(u=\frac{1}{2} \epsilon_0 E^2\)
The electric field of a point charge is:
\(E=\frac{k Q}{r^2}\)
Substitute \(E\) into the energy density equation:
\(u=\frac{1}{2} \epsilon_0\left(\frac{k Q}{r^2}\right)^2\)
\(
u=\frac{1}{2} \epsilon_0 \frac{k^2 Q^2}{r^4}
\)
Since \(\frac{1}{2} \epsilon_0 k^2 Q^2\) is constant, the energy density is proportional to \(\frac{1}{r^4}\) :
\(u \propto \frac{1}{r^4}\)

Hint

(b) Step 1: We have a parallel-plate capacitor with two plates of unequal area. The larger plate is connected to the positive terminal of a battery, while the smaller plate is connected to the negative terminal.
Hint: Visualize the capacitor setup and identify which plate is connected to which terminal of the battery.
Step 2: Identify the Charges
When a capacitor is connected to a battery, the plate connected to the positive terminal accumulates a positive charge (\((Q+)\)) and the plate connected to the negative terminal accumulates a negative charge (\((Q-)\)).
Hint: Remember that the convention is that the positive charge appears on the plate connected to the positive terminal.
Step 3: Relate Charge to Voltage
The relationship between the charge \((Q)\) on the plates of a capacitor and the voltage \((\mathrm{V})\) across the capacitor is given by the equation:
\(
Q=C \cdot V
[latex]
where C is the capacitance of the capacitor.
Hint: Capacitance (C) is a property of the capacitor and does not depend on the area of the plates in this context.
Step 4: Analyze the Capacitance
For a parallel-plate capacitor, the capacitance is given by:
[latex]
C=\frac{\varepsilon_0 \cdot A}{d}
\)
where \(\varepsilon_0\) is the permittivity of free space, \(\boldsymbol{A}\) is the area of the plates, and \(d\) is the distance between the plates. However, this does not affect the relationship between the charges on the plates.
Hint: Focus on the fact that the capacitance relates to the geometry of the plates but does not change the fundamental charge relationship.
Step 5: Conclude the Relationship Between Charges Since the charges on the plates of a capacitor are equal in magnitude but opposite in sign, we can state:
\(
|Q+|=|Q-|
\)
This means that the positive charge on the larger plate \((Q+)\) is equal in magnitude to the negative charge on the smaller plate (\((Q-)\)).
Hint: Remember that in a capacitor, the total charge is conserved, leading to the conclusion that the magnitudes of the charges are equal.
Thus, we conclude that:
\(
Q+=Q-
\)

Hint

(d) The capacitance of a parallel plate capacitor is:
\(C=\frac{\epsilon_0 A}{d}\)
where:
\(C\) is the capacitance,
\(\epsilon_0\) is the permittivity of free space,
\(A\) is the area of one of the plates,
\(d\) is the distance between the plates.
When the metal plate touches both plates, the distance between the plates effectively becomes zero.
\(d=0\)
Substituting \(d=0\) into the capacitance formula:
\(C^{\prime}=\frac{\epsilon_0 A}{0}\)
\(C^{\prime}=\infty\)
However, in a real circuit, this results in a short circuit.
The capacitor discharges instantly.
The capacitance is undefined or indeterminate.

Hint

(c)

Region AB shows the potential difference across capacitor \(C_1\) and region CD shows the potential difference across capacitor \(C_2\). Now, we can see from the graph that region \(A B\) is greater than region CD. Therefore, the potential difference across capacitor \(C_1\) is greater than that across capacitor \(C_2\). \(\because\) Capacitance, \(\mathrm{C}=\frac{Q}{V}\)
\(\therefore \mathrm{C}_1<\mathrm{C}_2 \quad\) ( \(Q\) remains the same in series connection)

Hint

(a) increase.
Explanation:
When the plates are dipped in oil, the oil acts as a dielectric material between the plates. This dielectric material reduces the electric field between the plates compared to the air. The electric field is inversely proportional to the dielectric constant of the material between the plates. Since the dielectric constant of oil is higher than that of air, the electric field decreases when the plates are dipped in oil.

Oil between the plates of the capacitor acts as a dielectric. We know that the electric field decreases by a factor of \(\frac{1}{K}\) of the original field when we insert a dielectric between the plates of a capacitor (\(K\) is the dielectric constant of the dielectric). So, if the oil is pumped out, the electric field between the plates will increase, as the dielectric has been removed.

Hint

(b) When the spheres are connected, charges flow between them until they both acquire the same common potential \(V\).
The final charges on the spheres are given by
\(
\begin{aligned}
& \mathrm{Q}_1=\mathrm{C}_1 \mathrm{~V} \text { and } \mathrm{Q}_2=\mathrm{C}_2 \mathrm{~V} \\
& \therefore \frac{Q_1}{Q_2}=\frac{C_1 V}{C_2 V}=\frac{C_1}{C_2}
\end{aligned}
\)

Hint

(d) The formula for the equivalent capacitance of capacitors in series is:
\(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)
Since all capacitors are equal, \(C_1=C_2=C_3=C=6 \mu \mathrm{~F}\) :
\(\frac{1}{C_s}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\)
Therefore, the equivalent capacitance in series is:
\(C_s=2 \mu \mathrm{~F}\)
The formula for the equivalent capacitance of capacitors in parallel is:
\(C_p=C_1+C_2+C_3\)
Since all capacitors are equal, \(C_1=C_2=C_3=C=6 \mu \mathrm{~F}\) :
\(C_p=6+6+6=18 \mu \mathrm{~F}\)

Hint

(c) The capacitance of a capacitor is given by \(C=\frac{\epsilon_0 A}{d}\)
Here, \(A\) is the area of the plates of the capacitor and \(d\) is the distance between the plates.
So, we can clearly see that the capacitance of a capacitor does depend on the size and shape of the plates and the separation between the plates; it does not depend on the charges on the plates.

Hint

(b) When a dielectric slab is inserted between the plates of an isolated charged capacitor, the charge on the capacitor remains the same.

Explanation:

We know, conservation of charge states that total electric charge in an isolated system never changes. Here, the capacitor is kept isolated from the surroundings, hence charge cannot change by any means. So, the total charge of the system will be still the same as before inserting the dielectric slab.

Capacitance increases:
The capacitance of the capacitor increases when a dielectric is added (capacitor increases \(K\) times). This means that for the same amount of charge, the potential difference across the plates can decrease.

Electric field decreases:
When we insert a dielectric between the plates of a capacitor, induced charges of opposite polarity appear on the face of the dielectric. They build an electric field inside the dielectric, directed opposite to the original field of the capacitor. Thus, the net effect is a reduced electric field. The electric field potential difference of capacitor reduce to \(\frac{1}{K}[latex] times.

Potential difference decreases:
We know that the potential difference of a parallel plate charged capacitor is given by [latex]V=\frac{Q d}{\varepsilon A}\), where \(A\) is the area of the each parallel plate.
Since, total charge, the distance between the plates \(d\) and area \(A\) is constant then only changing the dielectric constant will change the potential difference.
Hence, inserting the dielectric slab will change the potential difference between the plates.

Stored energy decreases:
The stored energy in the capacitor decreases because the potential difference decreases, even though the charge remains the same.
We know that the electrostatic energy stored in a parallel plate capacitor with charge \(Q[latex] is given by, [latex]E=\frac{1}{2} C V^2\), where \(V\) is the potential difference between the plates and \(C\) is the capacitance. Since, the capacitance is given by \(C=\frac{\varepsilon A}{d}\),changes with changing the dielectric constant. Also the potential difference between the plates changes, if we put the value of these two we get the energy term as,
\(
E=\frac{1}{2} \frac{\varepsilon A}{d}\left(\frac{Q d}{\varepsilon A}\right)^2
\)
\(\Rightarrow E=\frac{1}{2} \frac{Q^2 d}{\varepsilon A}\), Which changes with the change in the dielectric constant.

Note: Here K is the dielectric constant of dielectric.

Hint

(d) We have a parallel plate capacitor with charge \(Q\) on one plate and \(-Q\) on the other plate.
When a dielectric slab with dielectric constant \(K\) is inserted, it affects the electric field between the plates.
Electric Field Without Dielectric:
The electric field \(E_0\) in the absence of the dielectric is given by:
\(
E_0=\frac{Q}{A \epsilon_0}
\)
where \(A\) is the area of the plates and \(\epsilon_0\) is the permittivity of free space.
Electric Field With Dielectric:
When the dielectric is inserted, the electric field \(E\) becomes:
\(
E=\frac{Q}{A \epsilon_0 K}
\)
This is because the dielectric reduces the electric field by a factor of \(K\).
Induced Charge on the Dielectric:
The dielectric slab will have induced charges on its surfaces due to the electric field. The charge induced on the positive surface of the dielectric is \(Q^{\prime}\).
Relation Between Charges:
The induced charge \(Q^{\prime}\) can be related to the electric field due to polarization. The electric field due to the induced charge is:
\(
E_p=\frac{Q^{\prime}}{A \epsilon_0}
\)
The net electric field \(E\) between the plates can be expressed as:
\(
E=E_0-E_p
\)
Setting Up the Equation:
Substituting the expressions for \(E_0[latex] and [latex]E_p\) :
\(
\frac{Q}{A \epsilon_0 K}=\frac{Q}{A \epsilon_0}-\frac{Q^{\prime}}{A \epsilon_0}
\)
Multiplying through by \(A \epsilon_0\) to eliminate the denominators:
\(
\frac{Q}{K}=Q-Q^{\prime}
\)
Solving for Induced Charge \(Q^{\prime}\) :
Rearranging the equation gives:
\(
Q^{\prime}=Q-\frac{Q}{K}
\)
Factoring out \(Q\) :
\(
Q^{\prime}=Q\left(1-\frac{1}{K}\right)
\)
Analyzing the Result:
Since \(K>1\) (the dielectric constant is always greater than 1 ), the term \(\left(1-\frac{1}{K}\right)\) is positive but less than 1. Therefore:
\(
Q^{\prime}<Q
\)

Hint

(a, c, d) Each plate of a parallel plate capacitor has a charge \(q\) on it (before the battery is connected). The capacitor is now connected to a battery, which will add \(q{\prime}\) charge on +ve side of the plate. So total charge becomes \(q+q{\prime}\). Similarly, on the other plate it will be \(q-q{\prime}\). That means the total charge on the two plates are not equal.

Hint

(b, c) Because the charge always remains conserved in an isolated system, it will remain the same.

Charge and potential difference:
Since \(Q=C V\) (where \(Q\) is charge, \(C\) is capacitance, and \(V\) is potential difference), if the capacitance decreases while the charge remains constant (because the capacitor is disconnected from the battery), then the potential difference must increase to maintain the same charge.

Energy:
The energy stored in a capacitor is given by \(U=\frac{1}{2} C V^2\). Since capacitance decreases and potential difference increases when the plates are separated, the energy stored in the capacitor also increases.

Hint

(d) Step 1: Understand the Configuration
We have a parallel-plate capacitor connected to a battery. The capacitor has two plates separated by a distance \(d\). A metal sheet of negligible thickness is inserted between the plates, remaining parallel to them.
Hint: Visualize the capacitor and the metal sheet to understand how they interact.
Step 2: Analyze the Effect of the Metal Sheet
When a metal sheet is placed between the plates of the capacitor, it will affect the electric field and the charge distribution. The metal sheet will induce charges on its surfaces due to the electric field between the capacitor plates.
Hint: Remember that conductors in an electric field will always have charges induced on their surfaces.
Step 3: Charge Induction on the Metal Sheet
The metal sheet will have equal and opposite charges induced on its two faces. The face of the metal sheet that is closer to the positive plate of the capacitor will acquire a negative charge, while the face closer to the negative plate will acquire a positive charge.
Hint: Think about how charges redistribute themselves in response to an electric field.
Step 4: Effect on Capacitance
The introduction of the metal sheet does not change the potential difference across the capacitor plates because the battery maintains a constant voltage. The capacitance \(C\) of a capacitor is given by the formula:
\(
C=\frac{Q}{V}
\)
Since the potential difference \(V\) remains constant and the charge \(Q\) does not change, the capacitance \(C\) also remains
unchanged.
Hint: Recall that capacitance depends on the physical configuration and the dielectric properties of the materials between the plates.
Step 5: Effect on Charge
Since the battery is connected, it will maintain the same potential difference across the capacitor. Therefore, the total charge \(Q\) on the capacitor remains constant. The presence of the metal sheet does not allow the capacitor to store more charge.
Hint: Consider how the battery regulates the charge on the capacitor.
Step 6: Conclusion
Based on the analysis:
The capacitance does not increase.
The potential difference does not change.
The charge does not increase.
Equal and opposite charges appear on the two faces of the metal sheet.
Thus, the correct conclusion is that equal and opposite charges will appear on the two faces of the metal sheet.

Alternate:

The capacitance of the capacitor in which a dielectric slab of dielectric constant K , area A and thickness \(t\) is inserted between the plates of the capacitor of area \(A\) and separated by a distance \(d\) is given by \(C=\frac{\in_0 A}{(d-t)+\left(\frac{t}{K}\right)}\)
Since it is given that the thickness of the sheet is negligible, the above formula reduces to \(\mathrm{C}=\frac{\in_0 A}{d}\). In other words, there will not be any change in the electric field, potential or charge.

Hint

(b) (b,c,d) Justification of option

Hint

(a) Using the relation, \(W=q V_A\) where, \(V_A\) is the electric potential at point \(A\). We have, \(W=\left(4 \times 10^{-6}\right)\left(10^4\right)=4 \times 10^{-2} \mathrm{~J}\)

Hint

(a) The potential will be more on the smaller sphere.
Explanation: The potential \((\mathrm{V})\) of a sphere is given by the formula \(\mathrm{V}=\mathrm{kQ} / \mathrm{r}\), where \(k\) is Coulomb’s constant, \(Q\) is the charge, and \(r\) is the radius of the sphere. Since the charge \(({Q})\) is the same for both spheres, the potential is inversely proportional to the radius \((\mathrm{V} \propto 1 / \mathrm{r})\). Therefore, the smaller the radius, the higher the potential.

Hint

(c) Electric potential at a distance \(r\) due to point charge \(Q\) is \(V=\frac{k Q}{r}\) and electric field at same point is \(E=\frac{k Q}{r^2}\)
\(
\begin{array}{r}
\Rightarrow \quad E=\frac{k Q}{(k Q / V)^2}=\frac{V^2}{k Q}=\frac{\left(Q \times 10^{11}\right)^2}{k Q}=4 \pi \varepsilon_0 Q \times 10^{22} \mathrm{~V} / \mathrm{m} \\
\\
\left(\because k=\frac{1}{4 \pi \varepsilon_0}\right)
\end{array}
\)

Hint

(c) The electrical potential produced by the nucleus at the position of the electron,
\(
\begin{aligned}
V & =9 \times 10^9 \times \frac{q}{r} \\
& =9 \times 10^9 \times \frac{\left(+1.6 \times 10^{-19}\right)}{0.53 \times 10^{-10}}=27.2 \mathrm{~V}
\end{aligned}
\)

Hint

(b) Obviously, from charge configuration, at the centre electric field is non-zero. Potential at the centre due to charge \(2 q, V_{2 q}=\frac{2 q}{r}\) and potential due to \(-q\) charge, \(V_{-q}=-\frac{q}{r}\) (where, \(r=\) distance of centre point)
\(\therefore\) Total potential, \(V=V_{2 q}+V_{-q}+V_{-q}=0\)

Hint

(b) The electric field intensity \(E\) and electric potential \(V\) are related as \(E=-\frac{d V}{d r}\) and for \(V=\) constant, \(\frac{d V}{d r}=0\) This imply that electric field intensity, \(E=0\).

Hint

\(
\text { (c) Potential difference, } \Delta V=\frac{W}{q}=\frac{2}{20}=0.1 \mathrm{~V}
\)

Hint

\(
\begin{aligned}
&\text { (c) Potential at } A=\text { Potential due to }(+q) \text { charge }+ \text { Potential }\\
&=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{\sqrt{a^2+b^2}}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q)}{\sqrt{a^2+b^2}}=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
\text { (a) } V_A & =10^{10}\left[\frac{-5 \times 10^{-6}}{15 \times 10^{-2}}+\frac{2 \times 10^{-6}}{5 \times 10^{-2}}\right]=\frac{1}{15} \times 10^6 \mathrm{~V} \\
V_B & =10^{10}\left[\frac{2 \times 10^{-6}}{15 \times 10^{-2}}-\frac{5 \times 10^{-6}}{5 \times 10^{-2}}\right]=-\frac{13}{15} \times 10^6 \mathrm{~V} \\
\therefore W & =q\left(V_A-V_B\right)=3 \times 10^{-6}\left[\frac{1}{15} \times 10^6-\left(-\frac{13}{15} \times 10^6\right)\right]=2.8 \mathrm{~J}
\end{aligned}
\)

Hint

(c) Potential at \(P\) due to \((+q)\) charge
\(
V_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}
\)
Potential at \(P\) due to \((-q)\) charge, \(V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z+a)}\)
Total potential at \(P\) due \((A B)\) electric dipole,
\(
\begin{aligned}
V & =V_1+V_2 \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}-\frac{1}{4 \pi \varepsilon_0} \frac{q}{(z+a)} \\
& =\frac{2 q a}{4 \pi \varepsilon_0\left(z^2-a^2\right)}
\end{aligned}
\)

Hint

\(
\text { (b) Electric field, } E=\frac{V}{d}=\frac{10}{2 \times 10^{-2}}=500 \mathrm{~N} / \mathrm{C}
\)

Hint

\(
\begin{aligned}
&\text { (a) We have, } V=E \times d\\
&\Rightarrow \text { Distance, } d=\frac{V}{E}=\frac{3000}{500}=6 \mathrm{~m}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Potential at }\\
&\begin{aligned}
C, V & =2\left(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}\right) \\
& =\left(9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.2}\right) \times 2=36 \times 10^4 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

(a) The electric potential \(V(x, y, z)=4 x^2 \mathrm{~V}\)

Now, \(\mathbf{E}=-\left(\hat{\mathbf{i}} \frac{\partial V}{\partial x}+\hat{\mathbf{j}} \frac{\partial V}{\partial y}+\hat{\mathbf{k}} \frac{\partial V}{\partial z}\right)\)
Now, \(\frac{\partial V}{\partial x}=8 x, \frac{\partial V}{\partial y}=0\) and \(\frac{\partial V}{\partial z}=0\)
Hence, \(\mathbf{E}=-8 x \hat{\mathbf{i}} \mathrm{~V} / \mathrm{m}\). So, at point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\)
\(\mathbf{E}=-8 \hat{\mathbf{i}} \mathrm{~V} / \mathrm{m}\) or 8 , along negative \(X\)-axis.

Hint

\(
\begin{aligned}
&\text { (a) We have, electric field, }\\
&E=-\frac{d V}{d x}=-\frac{d}{d x}\left(5 x^2+10 x-9\right)=-10 x-10
\end{aligned}
\)
\(
\therefore \quad(E)_{x=1}=-10 \times 1-10=-20 \mathrm{~V} / \mathrm{m}
\)

Hint

(c) Work done in displacing a charge particle is given by \(W_{A B}=q\left(V_B-V_A\right)\) and the line integral of electrical field from point \(A\) to \(B\) gives potential difference \(V_B-V_A=-\int_A^B E \cdot d l\) For equipotential surface, \(V_B-V_A=0\) and hence \(W=0\).

Hint

(c) Equipotential line always perpendicular to electric field lines

We can see that three more points (1, 2 and 3) are perpendicular to the electric field apart from the shaded point.

Hint

\(
\begin{aligned}
&\text { (b) Potential, }\\
&\begin{aligned}
V=\frac{K q}{r}=\frac{K Z e}{r} & =\frac{9 \times 10^9 \times 50 \times 16 \times 10^{-19}}{9 \times 10^{-15}} \\
& =8 \times 10^6 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

(b) Since, potential inside the hollow sphere is same as that on the surface.

Hint

(a) Let \(Q_1\) and \(Q_2\) are the charges on sphere of radii \(R_1\) and \(R_2\), respectively.

Surface charge density, \(\sigma=\frac{\text { charge }}{\text { area }}\)
According to given problem, \(\sigma_1=\sigma_2\)
\(
\begin{aligned}
& \frac{Q_1}{4 \pi R_1^2} & =\frac{Q_2}{4 \pi R_2^2} \\
\therefore & \frac{Q_1}{Q_2} & =\frac{R_1^2}{R_2^2} \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
&\text { In case of a charged sphere, }\\
&V_s=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}
\end{aligned}
\)
\(
\begin{aligned}
V_1 & =\frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{R_1}, V_2=\frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{R_2} \\
\frac{V_1}{V_2} & =\frac{Q_1}{R_1} \times \frac{R_2}{Q_2}=\frac{Q_1}{Q_2} \times \frac{R_2}{R_1} \\
& =\left(\frac{R_1}{R_2}\right)^2 \times\left(\frac{R_2}{R_1}\right)=\frac{R_1}{R_2} \text { [using Eq. (i)] }
\end{aligned}
\)

Hint

(c) Let \(q^{\prime}\) be the charge on the inner shell. Then, the potential of the inner is \(V=\) potential due to its own \(q^{\prime}+\) potential due to the outer charge
\(
=\frac{1}{4 \pi \varepsilon}\left(\frac{q}{r_1}+\frac{q}{r_2}\right)
\)
But \(V=0\), because the inner shell is earthed.
\(
\therefore \quad \frac{q_1}{r_1}+\frac{q_2}{r_2}=0 \quad q^{\prime}=-q\left(\frac{r_1}{r_2}\right)
\)

Hint

(b) Since the point at 5 cm is inside the conductor, the potential at this point is equal to the potential at the surface.
Therefore, the potential at the surface is \(v\).
The potential at the surface of the sphere is given by \(V=\frac{k Q}{r}\), where \(r\) is the radius of the sphere.
Solving for \(Q[latex], we get [latex]Q=\frac{V r}{k}\).
Substituting \(r=10 \mathrm{~cm}=0.1 \mathrm{~m}\), we get \(Q=\frac{V \times 0.1}{k}\).
The potential at a distance \(r^{\prime}=15 \mathrm{~cm}=0.15 \mathrm{~m}\) from the center is given by \(V^{\prime}=\frac{k Q}{r^{\prime}}\). Substituting \(Q=\frac{0.1 V}{k}\), we get \(V^{\prime}=\frac{k}{0.15} \times \frac{0.1 V}{k}\).
Simplifying, we get \(V^{\prime}=\frac{0.1 V}{0.15}=\frac{1}{1.5} V=\frac{2}{3} V\).
The potential at a point distant 15 cm from the center is \(\frac{2}{3} V\).

Hint

(c) Inside a conducting body, potential is same everywhere and equals to the potential at its surface.

The electric potential \(\boldsymbol{V}\) on the surface of the sphere is given by:
\(V=\frac{1}{4 \pi \epsilon_0} \frac{q}{R}\)

Since the sphere is a conductor, the electric potential inside is constant and equal to the potential on the surface.
Therefore, the electric potential at \(r=\frac{R}{3}\) is the same as on the surface.
\(V_{\text {inside }}=V_{\text {surface }}=\frac{1}{4 \pi \epsilon_0} \frac{q}{R}\)
The electric potential within the sphere at a distance \(r=\frac{R}{3}\) from its center is \(\frac{1}{4 \pi \epsilon_0} \frac{q}{R}\).

Hint

(d) If charge acquired by the smaller sphere is \(q\), then its potential, \(120=\frac{k q}{2} \dots(i)\).
Whole charge comes to outer sphere.
Also, potential of the outer sphere,
\(
V=\frac{k q}{6} \dots(ii)
\)
From Eq. (i) and (ii), we get
\(
V=40 \mathrm{~V}
\)

Hint

\(
\text { (a) } V_A=\frac{\sigma}{\varepsilon_0}(a-b+c)
\)

\(
\begin{aligned}
&\begin{aligned}
V_B & =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-b+c\right) \\
\Rightarrow \quad V_C & =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-\frac{b^2}{c}+c\right)
\end{aligned}\\
&\text { On putting } c=a+b \Rightarrow V_A=V_C \neq V_B
\end{aligned}
\)

Hint

(a) If charge on a conducting sphere of radius \(R\) is \(Q\) then potential outside the sphere,
\(
V_{\text {out }}=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}
\)
At the surface of sphere
\(
V_s=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}
\)

Hint

(c) Potential, \(V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}=9 \times 10^9 \times \frac{500 \times 10^{-6}}{1.0}\) \(=4.5 \times 10^6 \mathrm{~V}\)

Hint

(a) The kinetic energy gained by a charge of \(q\) coulomb through a potential difference of \(V\) volt is
\(
\begin{aligned}
& K=q V \mathrm{~J} \\
& K=\left(1.6 \times 10^{-19}\right)(2000)=3.2 \times 10^{-16} \mathrm{~J}
\end{aligned}
\)
We have,
\(
\begin{aligned}
& K=\frac{1}{2} m v^2 \\
& v=\sqrt{\frac{2 K}{m}}=\frac{2 \times 3.2 \times 10^{-16}}{9 \times 10^{-31}}=\frac{8}{3} \times 10^7 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Speed of the particle, } v=\sqrt{\frac{2 Q V}{m}}\\
&\Rightarrow \quad v \propto \sqrt{Q} \Rightarrow \frac{v_A}{v_B}=\sqrt{\frac{Q_A}{Q_B}}=\sqrt{\frac{16 q}{4 q}}=\frac{2}{1}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) The kinetic energy of the electron, }\\
&K=e V=\frac{1}{2} m v^2
\end{aligned}
\)
\(
V=\frac{1}{2} \frac{m v^2}{e}
\)
\(
\text { Here, } \quad v=\frac{c}{10}=\frac{3 \times 10^8}{10}=3 \times 10^7 \mathrm{~m} / \mathrm{s}
\)
\(
V=\frac{1}{2} \frac{\left(9.0 \times 10^{-31}\right)\left(3.0 \times 10^{-7}\right)^2}{\left(1.6 \times 10^{-19}\right)}
\)
\(
\text { Potential, } \quad V=2531 \mathrm{~V}
\)

Hint

(b) \(\Delta \mathrm{PE}=\) Work done by external agent \(=\left(V_f q-V_i q\right), V_f>V_i \Rightarrow \Delta \mathrm{PE}>0\), i.e. PE will increase.

Hint

(b) Potential energy of the system will be given by
\(
=\frac{(-e)(-e)}{4 \pi \varepsilon_0 r}=\frac{e^2}{4 \pi \varepsilon_0 r}
\)
As, \(r\) decreases, potential energy increases.

Hint

\(
\begin{aligned}
& \text { (c) Work done, } W=U_f-U_i=9 \times 10^9 \times Q_1 Q_2\left[\frac{1}{r_2}-\frac{1}{r_1}\right] \\
& \begin{aligned}
& \Rightarrow W=9 \times 10^9 \times 12 \times 10^{-6} \times 8 \times 10^{-6}\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{10 \times 10^{-2}}\right] \\
& \quad=12.96 \mathrm{~J} \approx 13 \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Hint

(a) Since, the proton is moving against the direction of electric field, so work is done by the proton against electric field. It implies that electric field does negative work on the proton. Again, proton is moving in electric field from low potential region to high potential region, hence its potential energy increases.

Hint

(c) Electric potential energy, \(U=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d}\)
\(
\begin{aligned}
\therefore \quad U & =\frac{\left(9 \times 10^9\right) \times\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right) \mathrm{J}}{10^{-10}} \\
& =-9 \times 10^9 \times 1.6 \times 10^{-9} \mathrm{eV}=-14.4 \mathrm{eV}
\end{aligned}
\)
Note In the solution given all the values are positive. It is important to mention the sign.

Hint

(c) Potential energy as well as force are positive, if there is repulsion between the particles and negative, if there is attraction.
We take only the magnitude of values when discussing decrease or increase of energy.
As, \(\quad U=\frac{Q_1 Q_2}{4 \pi \varepsilon_0 r}\)
Plus or minus i.e., whether both are of the same sign or different, if \(r\) decreases, the value increase. Therefore, option (c) is wrong.

Hint

(b) Potential energy of charges \(Q_1\) and \(Q_2\) at 10 cm apart,
\(
U_i=\frac{1}{4 \pi \varepsilon_0} \frac{12 \times 10^{-6} \times 5 \times 10^{-6}}{0.1}
\)
\(
=\frac{9 \times 10^9 \times 60 \times 10^{-12}}{0.1}=54 \times 10^{-1}=5.4 \mathrm{~J}
\)
Potential energy of charge \(Q_1\) and \(Q_2\) at 6 cm apart,
\(
U_2=\frac{9 \times 10^9 \times 60 \times 10^{-12}}{0.06}=9 \mathrm{~J}
\)
\(\therefore\) Work done \(=(9-5.4) \mathrm{J}=3.6 \mathrm{~J}\)

Hint

\(
\begin{gathered}
\text { (d) We have, } A B+A C=12 \mathrm{~cm} \dots(i) \\
A B \cdot A C=32 \mathrm{~cm}^2
\end{gathered}
\)
\(
\begin{aligned}
\therefore A B-A C & =\sqrt{(A B+A C)^2-4 A B \cdot A C} \\
& A B-A C \dots(ii)
\end{aligned}=4 \mathrm{~cm}
\)
From Eqs. (i) and (ii), we get
\(
A B=8 \mathrm{~cm}, A C=4 \mathrm{~cm}
\)
Potential energy at point \(A\),
\(
\begin{aligned}
U_A & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_A q_B}{A B}+\frac{q_A q_C}{A C}\right] \\
& =\frac{9 \times 10^9 \times 4 \times 10^{-12}}{10^{-2}}\left(\frac{1}{8}+\frac{1}{4}\right)=1.35 \mathrm{~J}
\end{aligned}
\)

Hint

(c) Potential energy, \(U=\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q_1 Q_2}{r_1}+\frac{Q_2 Q_3}{r_2}+\frac{Q_1 Q_3}{r_3}\right]\)
Net potential energy, \(U_{\text {net }}=\frac{3}{4 \pi \varepsilon_0} \cdot \frac{q^2}{l}\)

Hint

\(
\begin{aligned}
&\text { (a) Potential energy of the system, }\\
&\begin{gathered}
U=K \frac{Q q}{l}+\frac{K q^2}{l}+\frac{K q Q}{l}=0 \\
\Rightarrow \quad \frac{K q}{l}(Q+q+Q)=0 \Rightarrow Q=-\frac{q}{2}
\end{gathered}
\end{aligned}
\)

Hint

(d) Length of the diagonal of a cube having each side \(b\) is \(\sqrt{3} b\). So, the distance of centre of cube from each vertex is \(\frac{\sqrt{3} b}{2}\). Hence, the potential energy of the given system of charge is
\(
U=8 \times\left[\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(-q)(q)}{\sqrt{3} b / 2}\right]=\frac{-4 q^2}{\sqrt{3} \pi \varepsilon_0 b}
\)

Hint

\(
\text { (a) Change in potential energy }(\Delta U)=U_f-U_i
\)

\(
\begin{aligned}
&\text { Potential energy, }\\
&\begin{aligned}
\Delta U & =\frac{1}{4 \pi \varepsilon_0}\left[\left(\frac{q_1 q_3}{0.4}+\frac{q_2 q_3}{0.1}\right)-\left(\frac{q_1 q_3}{0.4}+\frac{q_2 q_3}{0.5}\right)\right] \\
\Delta U & =\frac{1}{4 \pi \varepsilon_0}\left[8 q_2 q_3\right]=\frac{q_3}{4 \pi \varepsilon_0}\left(8 q_2\right) \\
k & =8 q_2
\end{aligned}
\end{aligned}
\)

Hint

(c) Torque, \(\begin{aligned} \tau_{\max } & =p E=q(2 l) E=2 \times 10^{-6} \times 0.01 \times 5 \times 10^5 \\ & =10 \times 10^{-3} \mathrm{~N}-\mathrm{m}\end{aligned}\)

Hint

\(
\begin{gathered}
\text { (d) Work done, } W=p E(1-\cos \theta)=p E\left(1-\cos 180^{\circ}\right) \\
W=p E[1-(-1)]=2 p E
\end{gathered}
\)

Hint

(a) Given, \(q_1=1 \mathrm{C}, q_2=2 \mathrm{C}, q_3=3 \mathrm{C}\) and \(r_1=100 \mathrm{~cm}=1 \mathrm{~m}\) Initial PE of system, \(U_1=\frac{1}{4 \pi \varepsilon_0 r_1}\left(q_1 q_2+q_2 q_3+q_3 q_1\right)\)
\(
\begin{aligned}
& =\frac{9 \times 10^9}{1}(1 \times 2+2 \times 3+3 \times 1) \\
& =99 \times 10^9 \mathrm{~J}
\end{aligned}
\)
When \(r_2=50 \mathrm{~cm}=0.5 \mathrm{~m}\)
Final PE of system, \(U_2=\frac{1}{4 \pi \varepsilon_0 r_2}\left(q_1 q_2+q_2 q_3+q_3 q_1\right)\)
\(
=\frac{9 \times 10^9}{0.5}(1 \times 2+2 \times 3+3 \times 1)=2 \times 99 \times 10^9 \mathrm{~J}
\)
Work done,
\(
\begin{aligned}
W & =U_2-U_1=2 \times 99 \times 10^9-99 \times 10^9 \\
& =99 \times 10^9 \mathrm{~J}=9.9 \times 10^{10} \mathrm{~J}
\end{aligned}
\)

Hint

\(
\text { (d) } \mathrm{H}_2, \mathrm{O}_2, \mathrm{~N}_2 \text { etc., are not polar dielectrics. }
\)

Hint

(c) Volume of 8 small drops \(=\) Volume of big drop
\(
8 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \Rightarrow R=2 r
\)
As, capacity is proportional to \(r\), hence capacity become 2 times.

Hint

(c) The capacitance of an isolated spherical conductor is given by \(C=\frac{Q}{V}\), where \(Q\) is the charge on the sphere and \(V\) is the potential at the surface. The potential at the surface of a sphere with charge \(Q\) is given by \(V=\frac{Q}{4 \pi \epsilon_0 R}\). Therefore, the capacitance is \(C=\frac{Q}{Q /\left(4 \pi \epsilon_0 R\right)}=4 \pi \epsilon_0 R\).

Hint

(c) The capacitance of the Earth, modeled as a sphere, is given by \(C=4 \pi \varepsilon_0 r\), where \(r\) is the radius. Since the volume of a sphere is \(V=(4 / 3) \pi r^3\) and the surface area is \(A=4 \pi r^2\), we can express the radius in terms of volume and surface area.
From \(V=(4 / 3) \pi r^3\), we have \(r=\sqrt{\frac{3 V}{4 \pi}}\).
From \(A=4 \pi r^2\), we have \(r=\sqrt{\frac{A}{4 \pi}}\).
Using the relationship between surface area and volume, we can write:
\(
\begin{aligned}
& r^2=\frac{A}{4 \pi} \\
& r^3=\frac{3 V}{4 \pi}
\end{aligned}
\)
Dividing the second equation by the first, we get:
\(
\begin{aligned}
& \frac{r^3}{r^2}=\frac{\frac{3 V}{4 \pi}}{\frac{A}{4 \pi}} \\
& r=\frac{3 V}{A}
\end{aligned}
\)
Substituting this value of \(r\) into the capacitance equation:
\(
C=4 \pi \varepsilon_0 r=4 \pi \varepsilon_0 \frac{3 V}{A}=12 \pi \varepsilon_0 \frac{V}{A}
\)

Hint

(d) Given, \(2 \pi R=2 \Rightarrow R=\frac{1}{\pi}\)

For sphere \(C=4 \pi \varepsilon_0 K R\)
\(\Rightarrow \quad C=\frac{1}{9 \times 10^9} \times \frac{1}{\pi} \times 80 \quad\) (for water \(K=80[latex] )
Capacitance of sphere, [latex]C=282828 \mathrm{pF} \approx 2800 \mathrm{pF}\)

Hint

(d) We have, \(C=\frac{\varepsilon_0 A}{d} \Rightarrow C \propto \frac{1}{d}\)
Therefore, the capacitance of parallel plate condenser depends on the separation between the plates.

Hint

(c) Since, aluminium is a metal and very thin, therefore field inside this will be zero. Hence, it would not affect the field in between the two plates, so capacity \(=\frac{q}{V}=\frac{q}{E d}\), remains unchanged.

Explanation:

No change in field:
Since the aluminium sheet acts as a conductor, it doesn’t allow any electric field to penetrate through it, and thus the electric field between the plates of the capacitor remains the same as before the sheet was introduced.

No change in capacitance:
The capacitance of a capacitor is primarily determined by the geometric factors like the area of the plates and the distance between them, along with the dielectric constant of the material between the plates. Since the sheet doesn’t change these geometric parameters or the dielectric constant (air or vacuum is used in this context), the capacitance remains unchanged.

Hint

(a) The potential difference across the parallel plate capacitor,
\(
\begin{aligned}
V & =10-(-10)=20 \mathrm{volt} \\
\therefore \text { Capacitance } & =\frac{Q}{V}=\frac{40}{20}=2 \mathrm{~F}
\end{aligned}
\)

Hint

(d) \(C=\frac{\varepsilon_0 A}{d}\).
As \(A \rightarrow \frac{1}{2}\) times and \(d \rightarrow 2\) times So, \(C \rightarrow \frac{1}{4}[latex] times, i.e. [latex]C^{\prime}=\frac{1}{4} C=\frac{12}{4}=3 \mu \mathrm{~F}\)

Hint

\(
\begin{aligned}
& \text { (d) Capacitance, } C_{\text {medium }}=K C_{\text {air }} \\
& \Rightarrow \quad K=\frac{C_{\text {medium }}}{C_{\text {air }}}=\frac{110}{50}=2.20
\end{aligned}
\)

Hint

(d) Capacitance: \(C=500 \mu F=500 \times 10^{-6} F\)
Charging rate: \(\frac{d Q}{d t}=100 \mu \mathrm{C} / \mathrm{s}=100 \times 10^{-6} \mathrm{C} / \mathrm{s}\)
Target voltage: \(V=10 \mathrm{~V}\)
Use the formula \(Q=C V\).
Substitute the given values:
\(Q=\left(500 \times 10^{-6} F\right) \times(10 V)\)
\(Q=5000 \times 10^{-6} C\)
\(Q=5 \times 10^{-3} C\)
Use the formula \(t=\frac{Q}{\frac{d Q}{d t}}\).
Substitute the values:
\(t=\frac{5 \times 10^{-3} \mathrm{C}}{100 \times 10^{-6} \mathrm{C} / \mathrm{s}}\)
\(t=\frac{5 \times 10^{-3}}{10^{-4}} \mathrm{~s}\)
\(t=50 \mathrm{~s}\)
The time it takes for the capacitor to reach 10 V is 50 s.

Hint

\(
\begin{aligned}
& \text { (b) } C=\frac{\varepsilon_0 A}{d}=1 \mathrm{pF} \text { and } C=\frac{K \varepsilon_0 A}{2 d}=2 \mathrm{pF} \\
& \therefore K=4 .
\end{aligned}
\)

Hint

(b)

Capacitance in air, \(C_{\text {air }}=\frac{\varepsilon_0 A}{d}=9\)
Capacitance in medium,
\(
\begin{aligned}
\frac{1}{C_{\mathrm{med}}} & =\frac{1}{C_1}+\frac{1}{C_2}=\frac{d_1}{K_1 \varepsilon_0 A}+\frac{d}{K_2 \varepsilon_0 A} \\
\Rightarrow \quad C_{\mathrm{med}} & =\frac{K_1 K_2 \varepsilon_0 A}{K_1 d_2+K_2 d_1} \\
& =\frac{3 \times 6 \times \varepsilon_0 A}{3 \times 2 d / 3+6 \times d / 3}=\frac{18}{4} \times 9=40.5 \mathrm{pF}
\end{aligned}
\)

Hint

(a) Parallel plate capacitor, \(C=K \varepsilon_0 A / d\)
As, given in the figure, for series combination,
\(
\frac{1}{C^{\prime}}=\frac{1}{\frac{\varepsilon_0 A}{\frac{d}{2}}}+\frac{1}{\frac{2 \varepsilon_0 A}{\frac{d}{2}}} \Rightarrow {C^{\prime}}=\frac{4}{3} \frac{\varepsilon_0 A}{d}
\)

Hint

(c) Capacitor are in series, \(\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}\)
\(\therefore\) Capacitance, \(C^{\prime}=\frac{C}{3}\)
Total voltage of the series combination,
\(
V^{\prime}=V_1+V_2+V_3=V+V+V=3 V
\)

Hint

(c) In series, \(\frac{1}{C}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\) \(\Rightarrow\) Capacitance, \(C=\frac{2}{3} \mathrm{~F}\)

Hint

(d) Equivalent capacitance \(=\frac{2 \times 3}{2+3}=\frac{6}{5} \mu \mathrm{~F}\)
Total charge by \(Q=C V=\frac{6}{5} \times 1000=1200 \mu \mathrm{C}\)
Potential \((V)\) across \(2 \mu \mathrm{~F}\) is \(V=\frac{Q}{C}=\frac{1200}{2}=600 \mathrm{~V}\)
\(\therefore\) Potential on inner plates \(=1000-600=400 \mathrm{~V}\)

Hint

(c) Capacitors are in series, \(\frac{1}{C_{\mathrm{eq}}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{8}\)
\(
\Rightarrow \quad C_{\mathrm{eq}}=\frac{8}{13} \mu \mathrm{~F}
\)
Total charge, \(Q=C_{\mathrm{eq}} V=\frac{8}{13} \times 13=8 \mu \mathrm{C}\)
Potential difference across \(2 \mu \mathrm{~F}\) capacitor \(=\frac{8}{2}=4 \mathrm{~V}\)

Hint

(b) Capacitance, \(C_1=\frac{C}{4}\) (series)
Capacitance, \(C_2=4 C\) (parallel)
The ratio of capacitance,
\(
\therefore \quad \frac{C_1}{C_2}=\frac{C / 4}{4 C}=\frac{1}{16}
\)

Hint

(b) The given circuit can be drawn as follows

\(
\text { Effective capacitance, } C_{A B}=2+4=6 \mu \mathrm{~F}
\)

Hint

(a) 

\(
\begin{aligned}
&\text { Equivalent capacitance between } A \text { and } B \text { is }\\
&C_{A B}=\frac{C}{2}+\frac{C}{2}+C=2 C
\end{aligned}
\)

Hint

(c) The given circuit can be simplified as follows

\(
\text { Equivalent capacitance between } A \text { and } B \text { is } C_{A B}=2+2=4 \mu \mathrm{~F}
\)

Hint

(d) Potential difference across both the lines is same, i.e. 2 V . Hence, charge flowing in line (2).
\(
Q=\left(\frac{2 \times 2}{2+2}\right)=2 \mu \mathrm{C}
\)

So, charge on each capacitor in line (2) is \(2 \mu \mathrm{C}\) and charge in line (1) is \(Q=2 \times 1=2 \mu \mathrm{C}\).

Hint

(a)

Therefore, capacitor \(2 \mu \mathrm{~F}, 4 \mu \mathrm{~F}\) and \(2 \mu \mathrm{~F}\) are in parallel.
So, equivalent capacitance between \(A\) and \(B\)
\(
C_{A B}=2+4+2=8 \mu \mathrm{~F}
\)

Hint

(c) The energy stored in a capacitor is
\(
U=\frac{1}{2} C V^2
\)
So, potential difference, \(V=\sqrt{\frac{2 U}{C}}=\sqrt{\frac{2 \times 50}{100 \times 10^{-6}}}=1000 \mathrm{~V}\)

Hint

(a) In series capacitance, \(C^{\prime}=\left(\frac{C_1}{n_1}\right)\) and \(V^{\prime}=4 V\) Energy, \(U^{\prime}=\frac{1}{2} C^{\prime} V^{\prime 2}=\frac{1}{2}\left(\frac{C_1}{n_1}\right)(4 V)^2\)
In parallel capacitance, \(C^{\prime \prime}=n_2 C_2\) and \(V^{\prime \prime}=V\)
\(
U^{\prime \prime}=\frac{1}{2} C^{\prime \prime} V^{\prime \prime 2}=\frac{1}{2}\left(n_2 C_2\right) V^2
\)
Given, \(\frac{1}{2}\left(\frac{C_1}{n_1}\right)(4 V)^2=\frac{1}{2}\left(n_2 C_2\right) V^2\)
\(
\Rightarrow \quad C_2=\frac{16 C_1}{n_1 n_2}
\)

Hint

(b) Energy of capacitor, \(U=\frac{1}{2} \frac{Q^2}{C}\)
\(
\begin{array}{ll}
\therefore & 1.21 U=\frac{1}{2} \frac{(Q+2)^2}{C} \\
\therefore & \frac{1.21}{1}=\frac{(Q+2)^2}{Q^2} \\
\Rightarrow & \sqrt{\frac{1.21}{1}}=\frac{Q+2}{Q} \\
\Rightarrow & 1.1 Q=Q+2
\end{array}
\)
Charge on the capacitor, \(Q=20 \mathrm{C}\)

Hint

\(
\begin{aligned}
& \text { (a) Initial energy }=\frac{1}{2} \times 1 \times 10^{-6} \times(30)^2=450 \times 10^{-6} \mathrm{~J} \\
& \text { Final energy }=\frac{1}{2}\left(C_1+C_2\right) V^2 \quad\left(\because V=\frac{V_1 C_1+V_2 C_2}{C_1+C_2}\right)
\end{aligned}
\)
\(
\begin{gathered}
=\frac{1}{2} \times 3 \times 10^{-6} \times(10)^2=150 \times 10^{-6} \mathrm{~J} \\
\text { Loss of energy }=(450-150) \times 10^{-6} \mathrm{~J}=300 \times 10^{-6}=300 \mu \mathrm{~J}
\end{gathered}
\)

Hint

(d) Initial energy stored in the capacitor,
\(
U_i=\frac{1}{2} C V^2=\frac{1}{2} C \times(50)^2=\frac{1}{2} C(50)^2 \dots(i)
\)
After 2 s , when the potential drops by 10 V , the final potential is 40 V .
Final energy stored in the capacitor, \(U_f=\frac{1}{2} C(40)^2 \dots(ii)\)
Fraction of energy stored \(=\frac{U_f}{U_i}=\frac{\frac{1}{2} C(40)^2}{\frac{1}{2} C(50)^2}=\left(\frac{40}{50}\right)^2=0.64\)