NEET Practice Q & A

Hint

(c) Angle between equipotential surface and line of force is \(90^{\circ}\).

Hint

(a) Potential, \(V=E \cdot d\)
Distance from the point charge, \(d=\frac{V}{E}=\frac{3000}{500}=6 \mathrm{~m}\)

Hint

(a)

\(
\begin{aligned}
\text { Potential difference } & =\frac{\text { Work done }}{\text { Charge }} \\
W & =q V \\
10 & =5 \times V \\
V & =2 \mathrm{~V}
\end{aligned}
\)

Hint

(a) An electron entering a higher potential region from a lower potential region will increase its velocity.

Explanation:
Electrons are negatively charged particles. They naturally move towards regions of higher potential. When an electron enters a region of higher potential, it experiences an electric force that accelerates it in that direction. This acceleration increases the electron’s velocity.

Hint

(d) The electric potential energy, E , between the two charges, \(\mathrm{q}_1\) and \(\mathrm{q}_2\), separated by the distance, r , is given as \(E=K \frac{q_1 q_2}{r}\),
Where \({k}=\) constant
As the distance between the charges is increased, the energy will decrease if both the charges are of similar nature. But if the charges are oppositely charged, the energy will become less negative and, hence, will increase.

Hint

(a) If a positive charge is shifted from a low potential region to a high potential region, then the electric potential energy increases.

Explanation: A positive charge naturally wants to move towards a region of lower potential, so moving it to a higher potential region requires external work against the electric field, which results in an increase in potential energy.

Hint

(a) Work done, \(W_{A \rightarrow B}=U_B-U_A=q\left(V_B-V_A\right)\)
Potential difference,
\(
\begin{aligned}
V_B-V_A & =\frac{W_{A-B}}{q} \\
& =\frac{10 \times 10^{-3}}{5 \times 10^{-6}}=2000 \mathrm{~V}=2 \mathrm{kV}
\end{aligned}
\)

Hint

(c) The capacitance \(C\) of a parallel plate capacitor is given by the formula:
\(
C=\frac{\epsilon_0 A}{d}
\)
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.
Charge on the Plates (Q): The formula for capacitance does not include charge. The charge on the plates affects the voltage across the capacitor but does not affect the capacitance itself. Thus, capacitance does not depend on the charge on the plates.

Hint

(c) As the electric field outside the capacitor plates is zero and field only exists in the between the plates, the energy will reside in field between the plates.

Hint

(c) The electric dipole potential varies as \(1 / r\) at large distance, is not true, because \(V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{r^2}\), i.e. \(V \propto \frac{1}{r^2}\).

Hint

(b) The energy stored in a condenser (capacitor) is stored in the form of electrical potential energy. This energy is stored in the electric field between the capacitor plates.

\(
U=\frac{1}{2} Q V=\frac{Q^2}{2 C}=\frac{1}{2} C V^2
\)

Here’s a more detailed explanation:
Electrical Potential Energy:
The work done to move a charge in an electric field is stored as potential energy. When a capacitor is charged, the work done to move charges from one plate to the other is stored as electrical potential energy.

Hint

(a) \(U=\frac{Q^2}{2 C}\)
When a slab of dielectric constant \(K\) is inserted, then \(C^{\prime}=C K\)
\(
\begin{aligned}
& U^{\prime}=\frac{Q^2}{2 C^{\prime}}=\frac{Q^2}{2 C K} \\
& \Rightarrow \quad U^{\prime}=\frac{U_0}{K}
\end{aligned}
\)

Hint

(b) Potential at a point in a field is defined as the amount of work done in bringing a unit positive test charge \((q)\) from infinity to that point along any arbitrary path.
Potential, \(V=\frac{W}{q}\)
\(\therefore\) Work done, \(W=q V\)

Hint

(a) Force between the plates of a parallel plate capacitor is given by
\(
|F|=\frac{\sigma^2 A}{2 \varepsilon_0}=\frac{Q^2}{2 \varepsilon_0 A}=\frac{C V^2}{2 d}
\)

Hint

(c) The positively charged particle experiences electrostatic force along the direction of electric field, i.e. from high electrostatic potential to low electrostatic potential. As, the work is done by the electric field on the positive charge, hence electrostatic potential energy of the positive charge decreases.

Hint

(a) In this problem, the collection of charges, whose total sum is not zero, but with regard to great distance can be considered as a point charge. The equipotentials due to point charges are spherical in shape, as the electric potential due to point charge \(q\) is given by
\(
V=k \frac{q}{r}
\)
This suggest that, electric potentials 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

(a) Electron is moving in opposite direction of field, so field will produce an accelerating effect on electron.

Hint

\(
\begin{aligned}
&\text { (a) Capacitance of a metallic sphere, }\\
&\begin{aligned}
C & =4 \pi \varepsilon_0 r=1 \times 10^{-6} \\
\Rightarrow \quad r & =10^{-6} \times 9 \times 10^9 \\
& =9 \mathrm{~km}
\end{aligned}
\end{aligned}
\)

Hint

(b) Unit of \(E\) in SI system is \(E=\frac{F}{q_0}=\) newton/coulomb
As,
\(
E=-\frac{d V}{d r}
\)
So, unit of \(E\) is also volt/metre.
\(
\begin{array}{rlr}
q & =C V & \\
q & =C E d & (\because V=E d) \\
E & =\frac{q}{C d}=\frac{q V}{q d}=\frac{W}{q d} & \left(\because C=\frac{q}{V}\right) \\
& =\frac{\text { joule }}{\text { coulomb-metre }} &
\end{array}
\)
while J/C is the unit of electric potential.

Hint

(a) Since the electric potential is constant throughout the conductor, the potential at any point inside the sphere is the same as the potential on the surface.
The electric potential on the surface of the sphere is given by:
\(V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\)

Hint

(b) Radius of spherical conductor, \(R=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}\)
According to given situation, \(V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R}\)
\(
\begin{aligned}
V & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{5 \times 10^{-2}} \\
\Rightarrow \quad \frac{q}{4 \pi \varepsilon_0} & =5 \times 10^{-2} \mathrm{~V} \dots(i)
\end{aligned}
\)
Again, electric potential at distance \(r=30 \mathrm{~cm}=30 \times 10^{-2} \mathrm{~m}\) from the centre,
\(
\begin{aligned}
V^{\prime} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{30 \times 10^{-2}} \\
& =\left(\frac{q}{4 \pi \varepsilon_0}\right) \times \frac{1}{30 \times 10^{-2}} \\
& =\frac{5 \times 10^{-2}}{30 \times 10^{-2}} \mathrm{~V} \\
& =\frac{V}{6} \text { [from Eq. (i)] }
\end{aligned}
\)

Hint

(a) The potential difference is the absolute difference between the potentials of the two plates:
\(
\begin{aligned}
& \Delta V=\left|V_2-V_1\right| \\
& \Delta V=|30 \mathrm{~V}-(-10 \mathrm{~V})| \\
& \Delta V=40 \mathrm{~V}
\end{aligned}
\)
\(
\begin{aligned}
&\text { The electric field is the potential difference divided by the distance: }\\
&\begin{aligned}
E & =\frac{\Delta V}{d} \\
E & =\frac{40 \mathrm{~V}}{0.02 \mathrm{~m}} \\
E & =2000 \frac{\mathrm{~V}}{\mathrm{~m}}
\end{aligned}
\end{aligned}
\)

Hint

(d) Potential at a point due to electric dipole
\(
V=\frac{p \cos \theta}{r^2}
\)
If \(\theta=0^{\circ}\), then \(V\) will be maximum and if \(\theta=180^{\circ}\), then \(V\) will be minimum.

Hint

\(
\text { (c) } U=-p E \cos \theta, U \text { is minimum at } \theta=0^{\circ} \text {. }
\)

Hint

(a) Kinetic energy gained by \(\alpha\)-particle,

\(
\begin{aligned}
\mathrm{KE} & =q \cdot \Delta V=q\left(V_1-V_2\right)=2 e\left(V_1-V_2\right) \\
& =2 \times 1.6 \times 10^{-19}(70-50)=40 \mathrm{eV}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) }\\
&\begin{aligned}
\mathrm{KE}=\frac{1}{2} m v^2 & =\frac{1}{2} m(a t)^2=\frac{1}{2} m\left(\frac{q E}{m}\right)^2 t^2 \quad\left(\because a=\frac{q E}{m}\right) \\
& =\frac{E^2 q^2 t^2}{2 m}
\end{aligned}
\end{aligned}
\)

Hint

(a) Charge \(q_1=12 \mu \mathrm{C}=12 \times 10^{-6} \mathrm{C}\)
Charge \(q_2=8 \mu \mathrm{C}=8 \times 10^{-6} \mathrm{C}\)
Initial distance \(r_i=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Final distance \(r_f=10 \mathrm{~cm}-4 \mathrm{~cm}=6 \mathrm{~cm}=0.06 \mathrm{~m}\)
\(
\begin{aligned}
& \text { Work done, } W=U_f-U_i=\frac{q_1 q_2}{4 \pi \varepsilon_0}\left[\frac{1}{r_f}-\frac{1}{r_i}\right] \\
& \quad=\left(9 \times 10^9\right)\left(10^{-12}\right)(12)(8)\left[\frac{1}{0.06}-\frac{1}{0.1}\right]=5.8 \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Capacitance, }\\
&\begin{aligned}
C & =4 \pi \varepsilon_0 R \\
& =\frac{1}{9 \times 10^9} \times 6408 \times 10^3 \\
& =712 \mu \mathrm{~F}
\end{aligned}
\end{aligned}
\)

Hint

(d) The initial capacitance of the parallel plate capacitor is \(\boldsymbol{C}\).
A thin metal plate is inserted such that it touches both plates.
The capacitance of a parallel plate capacitor is given by \(C=\frac{\epsilon_0 A}{d}\), where \(A\) is the area of the plates, \(d\) is the distance between the plates, and \(\epsilon_0\) is the permittivity of free space.
When a conductor touches both plates of a capacitor, it effectively shorts the capacitor.
Since the metal plate touches both plates, the effective distance between the plates becomes zero: \(d^{\prime}=0\).
The new capacitance \(C^{\prime}\) is given by \(C^{\prime}=\frac{\epsilon_0 A}{d^{\prime}}\).
Substituting \(d^{\prime}=0\), we get \(C^{\prime}=\frac{\epsilon_0 A}{0}\).
Division by zero results in infinity, so \(C^{\prime}=\infty\).

Hint

(c) Energy stored, \(W=\frac{Q^2}{2 C}\)
\(\therefore \quad W \propto Q^2\)
(if \(C\) is constant)
\(\frac{W}{W^{\prime}}=\frac{Q^2}{(2 Q)^2}=\frac{Q^2}{4 Q^2}\)
\(\therefore \quad W^{\prime}=4 W\)

Hint

(c) The energy stored in a capacitor is given by \(E=\frac{1}{2} C V^2\).
When the capacitor is discharged, all the stored energy is converted into heat.
\(
\begin{aligned}
&\text { Heat produced in capacitor }=\text { Energy of charged capacitor }\\
&=\frac{1}{2} C V^2=\frac{1}{2} \times\left(2 \times 10^{-6}\right) \times(100)^2=0.01 \mathrm{~J}
\end{aligned}
\)

Hint

(b) The equivalent capacitance \(C_{e q}\) of \(n\) capacitors in parallel is the sum of their individual capacitances:
\(C_{e q}=C+C+\ldots+C=n C\)
The total energy \(\boldsymbol{U}\) stored in the equivalent capacitor is:
\(U=\frac{1}{2} C_{e q} V^2\)
Substituting \(C_{e q}=n C\)
\(U=\frac{1}{2}(n C) V^2\)
\(U=\frac{1}{2} n C V^2\)

Hint

\(
\begin{aligned}
&\text { (d) Change in energy of condenser, }\\
&\begin{aligned}
\Delta U & =U_2-U_1=\frac{1}{2} C_2 V^2-\frac{1}{2} C_1 V^2 \\
& =\frac{V^2}{2}\left(C_2-C_1\right)=\frac{(100)^2}{2}\left(10 \times 10^{-6}-2 \times 10^{-6}\right) \\
& =4 \times 10^{-2} \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Hint

(b) When capacitors are connected in series, the charge on each capacitor is the same.
The energy stored in a capacitor is given by \(U=\frac{Q^2}{2 C}, U \propto \frac{1}{C} \quad\left(\because Q_1=Q_2\right)\), where \(Q\) is the charge and \(C\) is the capacitance.
The ratio of energies stored in the two capacitors is:
\(\frac{U_1}{U_2}=\frac{\frac{Q^2}{2 C_1}}{\frac{Q^2}{2 C_2}}\)
\(\frac{U_1}{U_2}=\frac{C_2}{C_1}\)
Substitute \(C_1=0.3 \mu \mathrm{~F}\) and \(C_2=0.6 \mu \mathrm{~F}\) into the ratio:
\(\frac{U_1}{U_2}=\frac{0.6 \mu \mathrm{~F}}{0.3 \mu \mathrm{~F}}\)
\(\frac{U_1}{U_2}=\frac{0.6}{0.3}\)
\(\frac{U_1}{U_2}=2\)

Hint

(b) When capacitors are connected in parallel, the voltage across them is the same.
The total charge in an isolated system remains constant.
The charge on a capacitor is given by \(Q=C V\), where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the voltage.
The initial charge on \(C_1\) is given by:
\(Q_1=C_1 V_0\)
Since the battery is removed, the total charge is conserved:
\(Q_{\text {total }}=Q_1=C_1 V_0\)
The equivalent capacitance of two capacitors in parallel is the sum of their individual capacitances:
\(C_{e q}=C_1+C_2\)
The final potential difference \(V_f\) across the combination is given by:
\(V_f=\frac{Q_{\text {total }}}{C_{e q}}\)
\(V_f=\frac{C_1 V_0}{C_1+C_2}\)
The potential difference across the combination is \(\frac{C_1 V_0}{C_1+C_2}\).

Hint

(c) Let \(Q_0, C_0, V_0\) and \(U_0\) be the charge, capacitance, potential difference and stored energy, respectively before the dielectric slab of dielectric constant \(K\) is inserted.
After inserting dielectric slab, charge will remain same, i.e. \(Q^{\prime}=Q_0\)
Potential difference will decrease as \(V^{\prime}=\frac{V_0}{K}\) Stored energy will decrease as, \(U=\frac{V_0}{K}\).

Note: Battery in disconnected so \(Q\) will be constant as \(C \propto K[latex]. So with introduction of dielectric slab capacitance will increase using [latex]Q=C V, V\) will decrease and using \(U=\frac{Q^2}{2 C}\), energy will decrease.

Hint

(b) Potential on parallel plate capacitor, \(V=\frac{Q}{C}\).
The capacitance of a parallel plate capacitor is given by the formula \(C=\frac{\epsilon_0 A}{d}\), where \(\epsilon_0\) is the permittivity of free space, \(A\) is the plate area, and \(d\) is the distance between the plates.
If the distance between the plates \((d)\) increases while the charge remains constant, the capacitance ( \(C\) ) decreases. Since \(V=\frac{Q}{C}\), if \(Q\) is constant and \(C\) decreases, then \(V\) must increase.

Hint

(b) Potential difference \(=V_i-V_o=\frac{q_i}{4 \pi \varepsilon_0}\left(\frac{1}{r_i}-\frac{1}{r_0}\right)\)
If \(q_i\) is positive, \(V_i-V_0=\) positive or \(V_i>V_o\).

Explanation: The electric potential \(V\) at a distance \(r\) from a charged sphere is given by the formula:
\(
V=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r}
\)

Hint

(a) Dielectric constant,
\(
\begin{aligned}
K & =\frac{\text { Permittivity of medium }}{\text { Permittivity of free space }} \\
K & =\frac{\varepsilon}{\varepsilon_0}
\end{aligned}
\)
\(\therefore\) Permittivity of water, \(\varepsilon=K \varepsilon_0\)
\(
\begin{aligned}
& =81 \times 8.85 \times 10^{-12} \\
& =7.16 \times 10^{-10} \text { MKS units }
\end{aligned}
\)

Hint

(c) Both the conductors carry equal and opposite charges. So, after connecting by a wire, there will be no charge in any conductor. Hence, all the stored energy will be neutralised. Loss of energy \(=2\left(\frac{1}{2} C V^2\right)=C V^2\)

Alternate:
The initial energy of the first conductor is:
\(U_1=\frac{1}{2} C V^2\)
The initial energy of the second conductor is:
\(U_2=\frac{1}{2} C(-V)^2=\frac{1}{2} C V^2\)
The total initial energy is:
\(U_{\text {initial }}=U_1+U_2=\frac{1}{2} C V^2+\frac{1}{2} C V^2=C V^2\)
The initial charge on the first conductor is:
\(Q_1=C V\)
The initial charge on the second conductor is:
\(Q_2=C(-V)=-C V\)
The total initial charge is:
\(Q_{\text {total }}=Q_1+Q_2=C V-C V=0\)
After connecting, the total charge remains the same, so the final charge is also 0. The final potential \(V_{\text {final }}\) is:
\(V_{\text {final }}=\frac{Q_{\text {total }}}{2 C}=\frac{0}{2 C}=0\)
\(
\begin{aligned}
&\text { The final energy of the system is: }\\
&\text { – } U_{\text {final }}=\frac{1}{2}(2 C) V_{\text {final }}^2=\frac{1}{2}(2 C)(0)^2=0
\end{aligned}
\)
The loss of energy is:
\(\Delta U=U_{\text {initial }}-U_{\text {final }}=C V^2-0=C V^2\)

Hint

(d) Total charge \(\mathrm{Q}=80+40=120 \mu \mathrm{C}\).
By using the formula \(Q_1{ }^{\prime}=Q_1\left[\frac{r_1}{r_1+r_2}\right]\).
New charge on sphere A is \(\mathrm{Q}_{\mathrm{A}}^{\prime}=\mathrm{Q}\left[\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}}\right]=120\left[\frac{4}{4+6}\right]=48 \mu \mathrm{C}\).
Initially it was \(80 \mu \mathrm{C}\) i.e., \(32 \mu \mathrm{C}\) charge flows from A to B.

Alternate:
Radius of sphere A: \(r_A=4 \mathrm{~cm}\)
Radius of sphere B: \(r_B=6 \mathrm{~cm}\)
Charge on sphere A: \(Q_A=80 \mu \mathrm{C}\)
Charge on sphere B: \(Q_B=40 \mu \mathrm{C}\)
When two conductors are connected, they reach the same potential.
The potential of a sphere with charge \(Q\) and radius \(r\) is \(V=\frac{k Q}{r}\), where \(k\) is Coulomb’s constant.
The total charge is the sum of the charges on both spheres:
\(Q_{\text {total }}=Q_A+Q_B\)
\(Q_{\text {total }}=80 \mu \mathrm{C}+40 \mu \mathrm{C}=120 \mu \mathrm{C}\)
After connecting the spheres, their potentials are equal:
\(V_A=V_B\)
\(\frac{k Q_A^{\prime}}{r_A}=\frac{k Q_B^{\prime}}{r_B}\)
\(\frac{Q_A^{\prime}}{r_A}=\frac{Q_B^{\prime}}{r_B}\)

The total charge is conserved:
\(Q_A^{\prime}+Q_B^{\prime}=Q_{\text {total }}\)
Solve for \(Q_A^{\prime}\) and \(Q_B^{\prime}\) :
\(Q_B^{\prime}=Q_{\text {total }}-Q_A^{\prime}\)
\(\frac{Q_A^{\prime}}{r_A}=\frac{Q_{\text {total }}-Q_A^{\prime}}{r_B}\)
\(Q_A^{\prime} r_B=Q_{\text {total }} r_A-Q_A^{\prime} r_A\)
\(Q_A^{\prime}\left(r_A+r_B\right)=Q_{\text {total }} r_A\)
\(Q_A^{\prime}=\frac{Q_{\text {total }} r_A}{r_A+r_B}\)
\(Q_A^{\prime}=\frac{120 \mu \mathrm{C} \cdot 4 \mathrm{~cm}}{4 \mathrm{~cm}+6 \mathrm{~cm}}=\frac{480}{10} \mu \mathrm{C}=48 \mu \mathrm{C}\)
\(Q_B^{\prime}=Q_{\text {total }}-Q_A^{\prime}=120 \mu \mathrm{C}-48 \mu \mathrm{C}=72 \mu \mathrm{C}\)
The charge flow is the difference between the initial and final charges on sphere A:
\(\Delta Q=Q_A-Q_A^{\prime}\)
\(\Delta Q=80 \mu \mathrm{C}-48 \mu \mathrm{C}=32 \mu \mathrm{C}\)
Since \(Q_A\) decreased, the charge flowed from A to B.
The amount of charge flowing from sphere A to sphere B is \(32 \mu \mathrm{C}\).

Hint

\(
\text { (a) Force, } F_e=q E=q\left(\frac{V}{d}\right)=\left(4 \times 10^{-6}\right)\left(\frac{2000}{2 \times 10^{-3}}\right)=4 \mathrm{~N}
\)

Hint

(a)
The electric potential on the surface of a conducting sphere is given by \(V=\frac{q}{r}\), where \(q\) is the charge and \(r\) is the radius.
When two conductors are in contact, charge flows until their electric potentials are equal.
Charge flows from higher potential to lower potential.
\(
\begin{aligned}
&\text { Calculate the electric potential of sphere A. }\\
&\begin{aligned}
& V_A=\frac{q_A}{r_A} \\
& V_A=\frac{18 \times 10^{-8} \text { statcoulomb }}{4 \mathrm{~cm}} \\
& V_A=4.5 \times 10^{-8} \frac{\text { statcoulomb }}{\mathrm{cm}}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& V_B=\frac{q_B}{r_B} \\
& V_B=\frac{9 \times 10^{-8} \text { statcoulomb }}{2 \mathrm{~cm}} \\
& V_B=4.5 \times 10^{-8} \frac{\text { statcoulomb }}{\mathrm{cm}}
\end{aligned}
\)
\(
V_A=V_B=4.5 \times 10^{-8} \frac{\text { statcoulomb }}{\mathrm{cm}}
\)
Since the potentials are equal, there is no potential difference. Therefore, no charge will flow.

Short-cut:

\(
\begin{aligned}
&\begin{array}{ll}
\text { (a) } & \left(\frac{q}{R}\right)_1=\left(\frac{q}{R}\right)_2 \\
\therefore & V_1=V_2
\end{array}\\
&\text { So, the charge will not flow at all. }
\end{aligned}
\)

Hint

(d) a decrease in energy of the system unless \(Q_1 R_2=Q_2 R_1\).

Explanation:
When two charged spheres are connected, they will redistribute their charges until they reach the same electric potential. This redistribution typically leads to a decrease in the overall energy of the system. However, there is a specific condition where no change occurs: if the ratio of charge to radius for each sphere is equal, meaning \(Q_1 / R_1=Q_2 / R_2\) or \(Q_1 R_2=Q_2 R_1\). In this case, the spheres already have the same potential and connecting them does not cause any change in their energy.
Why other options are incorrect:
(a) An increase in the energy of the system:
This is incorrect because connecting charged spheres generally leads to a decrease in energy. The energy is redistributed, but the total energy of the system decreases unless the specific condition mentioned in (d) is met.
(b) No change in the energy of the system:
This is only true if \(Q_1 R_2=Q_2 R_1\). Otherwise, the energy will always decrease.
(c) Always decrease in energy:
This is almost correct, but it’s slightly off. The energy does always decrease unless the condition \(Q_1 R_2=Q_2 R_1\) is met.

Hint

(a) Potential of the smaller sphere: \(\boldsymbol{V}_1=50 \mathrm{~V}\)
Potential of the bigger sphere: \(V_2=100 \mathrm{~V}\)
Electricity flows from a higher potential to a lower potential.
When a charged conductor is placed inside another hollow charged conductor and connected by a wire, both conductors attain the same potential.
The potential inside a hollow charged sphere is the same as the potential on its surface.
Compare the potentials of the two spheres and determine the direction of charge flow when they are connected.
Analyze the potential difference
Since \(V_2>V_1\), the bigger sphere has a higher potential.
Electricity flows from higher potential to lower potential.
If the smaller sphere is placed inside the bigger one and connected by a wire, the charge will flow from the bigger sphere to the smaller sphere until both reach the same potential.
If the spheres are placed side by side and connected, the charge will flow from the bigger sphere to the smaller sphere until both reach the same potential.
Electricity will flow from the bigger sphere to the smaller sphere in both cases. Therefore, electricity will flow from the smaller sphere to the bigger one only if the smaller sphere is placed inside the bigger one.

Hint

\(
\text { (b) Work done, } W=\Delta U=U_f-U_i=(3 U)-U=2 U
\)

Explanation:
Two identical charges are placed at two corners of an equilateral triangle.
The potential energy of the system is \(U\).
The potential energy between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by \(U=k \frac{q_1 q_2}{r}\), where \(k=\frac{1}{4 \pi \varepsilon_0}\).
The work done in bringing a charge from infinity to a point is equal to the change in potential energy.
Calculate the initial potential energy \(U\)
The initial potential energy of the system with two charges \(q\) separated by a distance \(r\) is:
\(U=k \frac{q^2}{r}\)
When a third identical charge \(q\) is brought to the third vertex, there are three pairs of charges:
\(U_f=k \frac{q^2}{r}+k \frac{q^2}{r}+k \frac{q^2}{r}=3 k \frac{q^2}{r}\)
The work done in bringing the third charge from infinity is the change in potential energy:
\(W=U_f-U\)
\(W=3 k \frac{q^2}{r}-k \frac{q^2}{r}\)
\(W=2 k \frac{q^2}{r}\)
\(W=2 U\)
The work done in bringing an identical charge from infinity to the third vertex is \(2 U\).

Hint

\(
\text { (b) Potential, } V=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}
\)

Here,
\(
\begin{aligned}
& V=2 V_{+\mathrm{ve}}+2 V_{-\mathrm{ve}} \\
& V=\frac{1}{4 \pi \varepsilon_0}\left[\frac{2 q}{L}-\frac{2 q}{L \sqrt{5}}\right]
\end{aligned}
\)
The electric potential at mid-point \(A\),
\(
V=\frac{2 q}{4 \pi \varepsilon_0 L}\left(1-\frac{1}{\sqrt{5}}\right)
\)

Hint

(a) The electric potential inside a hollow conducting sphere is constant and equal to the potential on its surface.
The potential at the center of the sphere is equal to the potential on its surface.
State the potential at the center:
The potential at the center of the sphere is the same as the potential on the surface.
\(
V_c=V_s
\)
The potential at the center of the sphere is 80 V.

Hint

\(
\begin{aligned}
&\text { (c) The energy stored in the capacitor, }\\
&\begin{aligned}
& U=\frac{1}{2} C V^2 \\
& U=\frac{1}{2}\left(\frac{A \varepsilon_0}{d}\right)(E d)^2 \quad\left(\because C=\frac{A \varepsilon_0}{d} \text { and } V=E d\right) \\
& U=\frac{1}{2} \varepsilon_0 E^2 A d
\end{aligned}
\end{aligned}
\)

Hint

(d) Given, \(q_1=5 \mu \mathrm{C}, q_2=10 \mu \mathrm{C}\) and \(r=0.5 \mathrm{~m}\)
The work done is equal to the change in potential energy.
The potential energy between two point charges is given by \(U=k \frac{q_1 q_2}{r}\).
Work done,
\(
\begin{aligned}
W & =K \frac{q_1 q_2}{r} \\
& =9 \times 10^9 \times \frac{5 \times 10^{-6} \times 10 \times 10^{-6}}{0.5} \\
& =9 \times 10^{-1} \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Here, }\\
&\begin{aligned}
& q=4 \times 10^{-3} \mathrm{C}, E=5 \mathrm{~V} / \mathrm{m} \\
& t=10 \mathrm{~s}, m=2 \times 10^{-3} \mathrm{~kg}
\end{aligned}\\
&\begin{aligned}
\text { The KE of charged particle } & =\frac{q^2 E^2 t^2}{2 m} \\
& =\frac{\left(4 \times 10^{-3}\right)^2 \times(5)^2 \times(10)^2}{2 \times 2 \times 10^{-3}}
\end{aligned}\\
&\therefore \quad \mathrm{KE}=10 \mathrm{~J}
\end{aligned}
\)

Hint

(d) The potential which is required to ionise the electron from outermost sheel of mercury is called ionisation potential. The electric field strength is given by \(E=\frac{V}{d}\). where, \(d\) is distance between plates creating electric field.
Given,
\(
V=10.39 \mathrm{~V}, E=1.5 \times 10^6 \mathrm{~V} / \mathrm{m}
\)
Distance travelled by electron to gain ionisation energy,
\(
\therefore \quad d=\frac{V}{E}=\frac{10.39}{1.5 \times 10^6} \mathrm{~m}
\)

Hint

\(
\begin{aligned}
&\text { (a) By using charge conservation, }\\
&\begin{aligned}
0.2 \times 600 & =(0.2+1) V \\
V & =\frac{0.2 \times 600}{1.2}=100 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Note: \(\text { The charge on a capacitor is given by } Q=C V \text {. }\)

Hint

(d) Here, \(q=8 \times 10^{-18} \mathrm{C}, C=100 \mu \mathrm{~F}=10^{-4} \mathrm{~F}\) Potential, \(V=\frac{q}{C}=\frac{8 \times 10^{-18}}{10^{-4}}=8 \times 10^{-14} \mathrm{~V}\)
\(
\begin{aligned}
\text { Work done } & =\frac{1}{2} q V=\frac{1}{2} \times 8 \times 10^{-18} \times 8 \times 10^{-14} \\
& =32 \times 10^{-32} \mathrm{~J}
\end{aligned}
\)

Hint

(c) Capacitance of a parallel plate capacitor with air: \(C=\frac{\epsilon_0 A}{d}\)
Capacitance with a dielectric material: \(C^{\prime}=\frac{\epsilon_0 A}{d-t+\frac{t}{k}}\)
The initial capacitance is given by:
\(C_1=\frac{\epsilon_0 A}{d_1}\)
\(C_1=\frac{\epsilon_0 A}{3}\)
The capacitance with the dielectric is given by:
\(C_2=\frac{\epsilon_0 A}{d_2-t+\frac{t}{k}}\)
\(C_2=\frac{\epsilon_0 A}{d_2-1+\frac{1}{2}}\)
\(C_2=\frac{\epsilon_0 A}{d_2-\frac{1}{2}}\)
Since the capacitance must remain the same, \(C_1=C_2\) :
\(\frac{\epsilon_0 A}{3}=\frac{\epsilon_0 A}{d_2-\frac{1}{2}}\)
\(3=d_2-\frac{1}{2}\)
\(d_2=3+\frac{1}{2}\)
\(d_2=\frac{7}{2}\)
\(d_2=3.5 \mathrm{~mm}\)
The new separation between the plates must be 3.5 mm.

Hint

(d) Electric field, \(E=-\frac{d V}{d r}=-\frac{d}{d x}\left(3 x^2\right)=-6 x\)
At point \((2,0,1)\)
Electric field, \(E=-12 \mathrm{Vm}^{-1}\)

Hint

(c) Kinetic energy, \(K=\frac{1}{2} m v^2=e V\)
\(\Rightarrow\) Speed of the electron, \(v=\sqrt{\frac{2 e V}{m}}\).

Hint

\(
\begin{aligned}
&\text { (b) Kinetic energy, }\\
&=e\left(V_A-V_B\right)=1.6 \times 10^{-19}(70-50)=3.2 \times 10^{-18} \mathrm{~J}
\end{aligned}
\)

Hint

(c) By using, \(W=Q \cdot \Delta V\)
Potential difference,
\(
\Delta V=\frac{W}{Q}=\frac{2}{20}=0.1 \mathrm{~V}
\)

Hint

(b) The relationship between energy, charge, and potential difference is: \(E=Q V\)
Multiply the energy in eV by the conversion factor to get the energy in Joules:
\(E=4 \times 10^{20} \mathrm{eV} \times 1.602 \times 10^{-19} \frac{\mathrm{~J}}{\mathrm{eV}}\)
\(E=6.408 \times 10^1 \mathrm{~J}\)
\(E=64.08 \mathrm{~J}\)
Use the formula \(E=Q V\) and solve for \(V\) :
\(V=\frac{E}{Q}\)
\(V=\frac{64.08 \mathrm{~J}}{0.25 \mathrm{C}}\)
\(V=256 \mathrm{~V}\)

Hint

\(
\text { (a) By using } \mathrm{KE}=Q V \Rightarrow \mathrm{KE}=1.6 \times 10^{-19} \times 100
\)
\(
=1.6 \times 10^{-17} \mathrm{~J}
\)

Hint

(d) In static situations, the potential at all points on a conductor is the same.
The surface of a conductor is an equipotential surface.
Apply the concept of equipotential surfaces on conductors.
Apply the property of conductors
As the sphere is a conductor, the potential at all points on its surface is the same.
This is because the surface of a conductor is an equipotential surface in a static electric field.
Determine the potential at the three points
Since \(A, B\), and \(C\) all lie on the surface of the conductor, they all have the same potential.
Therefore, \(\mathrm{VA}=\mathrm{VB}=\mathrm{VC}\).

Hint

(a) Potential at \(O\) due to charge at \(A\),

\(
V_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{a}
\)
Potential at \(O\) due to charge at \(B\),
\(
V_2=\frac{1}{4 \pi \varepsilon_0} \frac{(-q)}{a}
\)
\(\therefore\) Potential at mid-point \(O\),
\(
V=\frac{1}{4 \pi \varepsilon_0} \frac{q}{a}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(-q)}{a}=0
\)

Hint

\(
\begin{aligned}
&\text { (b) As electric potential of spheres are same, }\\
&\begin{array}{rlrl}
& V_A & =V_B \\
\Rightarrow & E_A \cdot a & =E_B \cdot b \\
\therefore & \frac{\sigma_A a}{\varepsilon_0} & =\frac{\sigma_B b}{\varepsilon_0} \text { or } \frac{\sigma_A}{\sigma_B}=\frac{b}{a}
\end{array}
\end{aligned}
\)

Hint

\(
\begin{aligned}
\text { (a) Given, } C_1 & =2 \mu \mathrm{~F}, V_1=50 \mathrm{~V} \\
C_2 & =1 \mu \mathrm{~F}, \quad V_2=20 \mathrm{~V}
\end{aligned}
\)
Common potential,
\(
V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{(2 \times 50+1 \times 20) \times 10^{-6}}{(2+1) \times 10^{-6}}=40 \mathrm{~V}
\)
Loss of energy,
\(
\begin{aligned}
\Delta u & =\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2 \\
& =\frac{1}{2} \frac{2 \times 1}{(2+1)}(50-20)^2 \times 10^{-6} \\
& =300 \times 10^{-6}=300 \mu \mathrm{~J}
\end{aligned}
\)

Hint

(c) \(A, B, C, D, E\) lies on equipotential surface. On sphere, the surface potential is same. So,
\(
W_{A B}=W_{A C}=W_{A D}=W_{A E}=q\left(V_f-V_i\right)=\text { zero }
\)

Hint

(b) We have, \(E_0=-\frac{d V}{d x}\) or \(d V=-E_0 d x\)
On integrating both sides, we get
\(
\begin{aligned}
\int d V & =-\int E_0 d x \\
V_x & =-x E_0
\end{aligned}
\)

Hint

(a) Force, \(F=q E=q\left(\frac{V}{d}\right)\)
\(\therefore\) Potential difference between two points,
\(
V=\frac{F \cdot d}{q}=\frac{(3000)\left(10^{-2}\right)}{3}=10 \mathrm{~V}
\)

Hint

\(
\begin{aligned}
&\text { (b) When charge particle enters in a potential field, then }\\
&\begin{aligned}
& \frac{1}{2} m v^2=q V \\
\therefore & v=\sqrt{\left(\frac{2 q V}{m}\right)} \quad \therefore \frac{v_A}{v_B}=\sqrt{\frac{q}{4 q}}=\frac{1}{2}
\end{aligned}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
U_{A B} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(10)(10) \times 10^{-12}}{10 \times 10^{-2}} \\
U_{B C} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(10)(10) \times 10^{-12}}{10 \times 10^{-2}} \\
U_{A C} & =\frac{1}{4 \pi \varepsilon_0} \frac{(10)(10) \times 10^{-12}}{10 \times 10^{-2}} \\
U_{\text {total }} & =U_{A B}+U_{B C}+U_{C A} \\
& =\frac{3}{4 \pi \varepsilon_0}\left[\frac{100 \times 10^{-12}}{10 \times 10^{-2}}\right]=27 \mathrm{~J}
\end{aligned}
\)

Hint

(b) When charge particle enters in uniform electric field, then force on charged particle, \(F=q E\)
Also, \(F=m a\)
\(\therefore\) \(m a=q E\)
or acceleration of the particle,
\(
\begin{aligned}
a & =\frac{q E}{m}=\frac{3 \times 10^{-3} \times 80}{20 \times 10^{-3}} \\
& =12 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
So, from equations of motion,
\(
v=u+a t=20+12 \times 3=56 \mathrm{~m} / \mathrm{s}
\)

Hint

(a)

Potential,
\(
\begin{aligned}
& V_A=V_B=V_C=V_D=V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\
& V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\left(50 \times 10^{-6}\right)}{(\sqrt{2})}
\end{aligned}
\)
\(\therefore\) Potential at the centre of square,
\(
\begin{aligned}
V_0 & =4 V \\
V_0 & =4 \times\left[\frac{9 \times 10^9 \times 50 \times 10^{-6}}{\sqrt{2}}\right] \\
& =90 \sqrt{2} \times 10^4 \mathrm{~V}
\end{aligned}
\)
Work done in bringing a charge ( \(q=50 \mu \mathrm{C}\) ) from \(\infty\) to centre of the square is
\(
\therefore \quad W=q V_0=50 \times 10^{-6} \times 90 \sqrt{2} \times 10^4=63.64 \mathrm{~J} \simeq 64 \mathrm{~J}
\)

Hint

(c)

\(
\begin{aligned}
& U_{A B}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(q)(-2 q)}{a} \\
& U_{B C}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(-2 q)(q)}{a} \\
& U_{C A}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(q)(q)}{2 a}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Potantial energy of the system, }\\
&\therefore \quad U_{\text {system }}=\frac{1}{4 \pi \varepsilon_0} \cdot\left[\frac{-2 q^2}{a}-\frac{2 q^2}{a}+\frac{q^2}{2 a}\right]
\end{aligned}
\)
\(
U_{\text {system }}=\frac{-7 q^2}{8 \pi \varepsilon_0 a}
\)

Hint

(b) When \(\alpha\)-particle is accelerated through a potential difference \(V\), then kinetic energy of \(\alpha\)-particle (double positive charge)
\(
\begin{aligned}
K=q V & =\left(2 e \mathbb{V}=2 \times 1.6 \times 10^{-19} \times 10^6 \mathrm{~J}\right. \\
& =\frac{2 \times 1.6 \times 10^{-19} \times 10^6}{1.6 \times 10^{-19}} \mathrm{eV}=2 \mathrm{MeV}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) Since, we know that, } K=\frac{p^2}{2 m}\\
&\begin{array}{ll}
\text { Also, } & K=q V \\
\therefore & p=\sqrt{2 m q V}
\end{array}
\end{aligned}
\)
\(
\therefore \text { Ratio of momenta, } \frac{p_e}{p_\alpha}=\sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}}=\sqrt{\frac{m_e}{2 m_\alpha}} \quad\left(\because q_\alpha=2 q_e\right)
\)

Hint

(d) Velocity, \(v=a t=\left(\frac{q E}{m}\right) t \Rightarrow v \propto \frac{q}{m}\) or \(\mathrm{KE} \propto \frac{q^2}{m^2}\)
\(\therefore\) The ratio of kinetic energy, \(\frac{K_1}{K_2}=\left(\frac{2}{1 / 2}\right)^2=16: 1\)

Hint

(c) Capacity of spherical condenser, when outer sphere is earthed,

\(
\begin{aligned}
&C_1=4 \pi \varepsilon_0 \cdot \frac{a b}{b-a}\\
&\text { Capacity of spherical condenser when inner sphere is earthed }\\
&C_2=4 \pi \varepsilon_0 b+\frac{4 \pi \varepsilon_0 a b}{b-a}
\end{aligned}
\)

\(
\begin{aligned}
&C_1=4 \pi \varepsilon_0 \cdot \frac{a b}{b-a}\\
&\text { Capacity of spherical condenser when inner sphere is earthed }\\
&C_2=4 \pi \varepsilon_0 b+\frac{4 \pi \varepsilon_0 a b}{b-a}
\end{aligned}
\)

Hint

(d) Work done, \(W=U_f-U_i\)
Here, \(\quad U_f=0\)
\(
\begin{aligned}
\therefore \quad W & =-U_i=-\frac{1}{4 \pi \varepsilon_0}\left[\frac{q q_2}{r}+\frac{q_2 q_3}{r}+\frac{q_3 q_1}{r}\right] \\
& =-\frac{\left(9 \times 10^9\right)}{0.1}[(1)(-2)+(1)(4)+(-2)(4)] \times 10^{-12} \\
& =0.54 \mathrm{~J}
\end{aligned}
\)

Hint

(d)
\(
\begin{aligned}
& \text { Here, } \frac{1}{2} m v^2=\frac{q_1 q_2}{4 \pi \varepsilon_0}\left(\frac{1}{r_i}-\frac{1}{r_f}\right) \\
& \quad \frac{1}{2} \times 2 \times 10^{-3} \times v^2=\left(10^{-9}\right)\left(9 \times 10^9\right)
\end{aligned}
\)
Speed of particle,
\(
\begin{aligned}
v & =\sqrt{8.1 \times 10^3} \mathrm{~m} / \mathrm{s} \\
& =90 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

\(
\text { (b) Work done, } W=U_i-U_f=6\left[\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q \cdot q}{r}\right]-0=\frac{6 q^2}{4 \pi \varepsilon_0 r}
\)

Hint

(d) Given circuit is balanced Wheatstone bridge circuit.
For branch \(A C B, C^{\prime}=\frac{10 \times 10}{10+10}=5 \mu \mathrm{~F}\)

For branch \(A D B, C^{\prime \prime}=\frac{10 \times 10}{10+10}=5 \mu \mathrm{~F}\)
Their is no flow of charge in branch \(C D\).
So, equivalent capacitance between \(A B\),
\(
C_{A B}=C^{\prime}+C^{\prime \prime}=5+5=10 \mu \mathrm{~F}
\)

Hint

(a) When capacitors are connected in series, then
\(
\begin{aligned}
\frac{1}{C_S} & =\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{3}+\frac{1}{9}+\frac{1}{18} \\
\frac{1}{C_S} & =\frac{1}{2} \\
\Rightarrow \quad C_S & =2 \mu \mathrm{~F}
\end{aligned}
\)
When capacitors are joined is parallel, then
\(
\begin{aligned}
& C_P=3+9+18=30 \mu \mathrm{~F} \\
\therefore & \frac{C_S}{C_P}=\frac{2}{30}=\frac{1}{15}
\end{aligned}
\)

Hint

(a) Capacitance between \(A\) and \(B\),
\(
\begin{aligned}
&C_{A B}=3+1=4 \mu \mathrm{~F}\\
&\text { Capacitance between } A \text { and } C \text {, }
\end{aligned}
\)
\(
C_{A C}=\frac{3}{2}+\frac{3}{2}=3 \mu \mathrm{~F}
\)
\(
\therefore \quad \frac{C_{A B}}{C_{A C}}=\frac{4}{3}
\)

Hint

(b) Given circuit can be simplified as follows

\(
\begin{aligned}
&\text { So, equivalent capacitance between } A \text { and } B \text {, }\\
&C_{A B}=1+1=2 \mu \mathrm{~F}
\end{aligned}
\)

Hint

(c) Given circuit can be simplified as,

\(
\begin{aligned}
&\text { So, net capacitance between } A B \text {, }\\
&C_{A B}=3+2=5 \mu \mathrm{~F}
\end{aligned}
\)

Hint

(c) Given circuit can be redrawn as follows,

\(
\begin{aligned}
&\text { Equivalent capacitance between } A \text { and } B \text {, }\\
&C_{A B}=\frac{2 C}{3}+C=\frac{5}{3} C
\end{aligned}
\)

Hint

(d) Let \(C_i\) be the individual capacitance of each capacitor.
When \(n\) identical capacitors with capacitance \(C_i\) are connected in series, the equivalent capacitance \(C_s\) is given by:
\(\frac{1}{C_s}=\frac{1}{C_i}+\frac{1}{C_i}+\ldots+\frac{1}{C_i}\) (n times)
\(\frac{1}{C_s}=\frac{n}{C_i}\)
\(C_s=\frac{C_i}{n}\)
Given that the equivalent capacitance in series is \(C\), and \(n=7\), we have:
\(C=\frac{C_i}{7}\)
\(C_i=7 C\)
When \(n\) capacitors with capacitance \(C_i\) are connected in parallel, the equivalent capacitance \(C_p\) is given by:
\(C_p=C_i+C_i+\ldots+C_i\) ( n times)
\(C_p=n C_i\)
Substitute \(C_i=7 C\) into the equation:
\(C_p=7(7 C)\)
\(C_p=49 \mathrm{C}\)

Hint

(b) Capacitance of the capacitor: \(C=16 \mu \mathrm{~F}=16 \times 10^{-6} \mathrm{~F}\)
Potential difference with glass slab: \(V^{\prime}=\frac{1}{8} V\) where \(V\) is the original potential difference
The capacitance of a capacitor with a dielectric material is given by \(C^{\prime}=\kappa C\), where \(\kappa\) is the dielectric constant and \(c\) is the capacitance without the dielectric.
The relationship between charge, capacitance, and potential difference is \(Q=C V\).
The charge on the capacitor remains constant when a dielectric is inserted.
Use the relationship between charge, capacitance, and potential difference to find the dielectric constant.
Relate the charge on the capacitor before and after inserting the glass slab
The charge remains constant:
\(Q=C V=C^{\prime} V^{\prime}\)
\(
C V=(\kappa C)\left(\frac{1}{8} V\right)
\)
\(
\begin{aligned}
& 1=\kappa \frac{1}{8} \\
& \kappa=8
\end{aligned}
\)
The dielectric constant of glass is 8.

Hint

(c) Capacitance of a parallel plate capacitor: \(C=\frac{e_0 A}{d}\)
Capacitance with a dielectric: \(C^{\prime}=k C\)
Calculate the initial capacitance \(C_1\).
\(
\begin{aligned}
& C_1=\frac{\epsilon_0 A}{d_1} \\
& 10^{-12}=\frac{\epsilon_0 A}{d_1}
\end{aligned}
\)
\(
\begin{aligned}
C_2 & =k \frac{\epsilon_0 A}{d_2} \\
C_2 & =k \frac{\epsilon_0 A}{2 d_1} \\
C_2 & =\frac{k}{2} \frac{\epsilon_0 A}{d_1}
\end{aligned}
\)
\(
\begin{aligned}
C_2 & =\frac{k}{2} C_1 \\
C_2 & =\frac{4}{2} \times 10^{-12} \\
C_2 & =2 \times 10^{-12}
\end{aligned}
\)
The new capacitance is \(2 \times 10^{-12} \mathrm{~F}\).

Hint

(a)

The effective capacity of the combination,
\(
\frac{2 C}{3}=4 \mu \mathrm{~F}
\)
\(\therefore\) Capacitance, \(C=6 \mu \mathrm{~F}\)

Hint

(d) By the given arrangement, two capacitors are formed such a way that they are in parallel.
\(
C_{\text {net }}=2 C=2\left(\frac{\varepsilon_0 A}{d}\right)
\)

Hint

(a) According to the question, \(8\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi R^3\)
\(\therefore \quad\) Radius, \(R=2 r\)
Capacitance,
\(
\begin{aligned}
C & =4 \pi \varepsilon_0 r \\
C^{\prime} & =4 \pi \varepsilon_0 2 r \Rightarrow C^{\prime}=2 C
\end{aligned}
\)
Therefore, the capacitance of the bigger drop as campared to each smaller drop is 2 times.

Hint

\(
\begin{aligned}
& \text { (a) Given, } C=10 \mu \mathrm{~F}=10^{-5} \mathrm{~F} \\
& \Rightarrow \quad \frac{\varepsilon_0 A}{D}=10^{-5} \mathrm{~F} \dots(i)
\end{aligned}
\)
Now, both capacitors are in parallel.
\(
\therefore \quad C^{\prime}=C_1+C_2=\frac{\varepsilon_0(A / 2)}{d}+\frac{K \varepsilon_0(A / 2)}{D}
\)
\(
\begin{aligned}
C^{\prime} & =\frac{\varepsilon_0 A}{d}\left(\frac{1}{2}+2\right)=\frac{5}{2} \frac{\varepsilon_0 A}{d}=\frac{5}{2} \times 10^{-5} \text { [from Eq. (i)] } \\
& =25 \mu \mathrm{~F}
\end{aligned}
\)

Hint

(d) Capacity of capacitor, \(C=\frac{\varepsilon_0 A}{d}\) \(\Rightarrow \quad C \propto \frac{1}{d}\)
\(\therefore\) Capacitance, \(C^{\prime}=2 C\)
\(\therefore\) Extra charge flow, \(q=(2 C V-C V)=C V\)
\(\therefore\) Work done, \(W=q V=(C V) V=C V^2\)

Hint

(b) Initial capacitance: \(C_0=25 \mu \mathrm{~F}\)
Plate separation: \(d\)
Thickness of metallic foil: \(t=\frac{2}{7} d\)
Capacitance of a parallel plate capacitor: \(C=\frac{\epsilon_0 A}{d}\), where \(A\) is the area of the plates, \(d\) is the separation, and \(\epsilon_0[latex] is the permittivity of free space.
When a metallic foil is inserted, the effective separation between the plates is reduced.
The new separation [latex]d^{\prime}\) is the original separation \(d\) minus the thickness of the foil \(t\) :
\(d^{\prime}=d-t\)
\(d^{\prime}=d-\frac{2}{7} d\)
\(d^{\prime}=\frac{5}{7} d\)
The new capacitance \(C^{\prime}\) is given by:
\(C^{\prime}=\frac{\epsilon_0 A}{d^{\prime}}\)
\(C^{\prime}=\frac{e_0 A}{\frac{5}{7} d}\)
\(C^{\prime}=\frac{7}{5} \frac{\epsilon_0 A}{d}\)
\(C^{\prime}=\frac{7}{5} c_0\)
Substitute \(C_0=25 \mu \mathrm{~F}\) into the equation:
\(C^{\prime}=\frac{7}{5}(25 \mu \mathrm{~F})\)
\(C^{\prime}=7 \cdot 5 \mu \mathrm{~F}\)
\(C^{\prime}=35 \mu \mathrm{~F}\)
The new capacitance is \(35 \mu \mathrm{~F}\).

Hint

(b) On introducing dielectric \(K\) in a parallel plate capacitor, its capacity becomes,
\(
\begin{array}{ll}
& C^{\prime}=K C_0 \\
\therefore & C^{\prime}=5 C_0
\end{array}
\)
Also, energy stored, \(W_0=\frac{q^2}{2 C_0}\)
\(
\begin{array}{ll}
\therefore & W^{\prime}=\frac{q^2}{2 C^{\prime}}=\frac{q^2}{2 \times 5 C_0} \\
\therefore & \frac{W_0}{W^{\prime}}=\frac{5}{1} \\
\therefore & W^{\prime}=\frac{W_0}{5}
\end{array}
\)

Hint

\(
\text { (b) } \frac{C^{\prime}}{C}=\frac{\varepsilon_0 A / d-t}{\varepsilon_0 A / d}=\frac{d}{d-t}=\frac{d}{d / 2}=2: 1
\)
\(
\left[\because \text { in question, thickness }(t) \text { is } b=\frac{d}{2}\right]
\)

Hint

(d) Capacitance with a dielectric: \(\boldsymbol{C}=\boldsymbol{K} C_0\), where \(\boldsymbol{C}_0\) is the capacitance without the dielectric.
Relationship between charge, voltage, and capacitance: \(Q=C V\).
Relationship between electric field and voltage: \(E=\frac{\boldsymbol{V}}{\boldsymbol{d}}\), where \(d\) is the distance between the plates.
The new capacitance \(C_2\) is given by \(C_2=K_2 C_0\).
Since \(C_0=\frac{C_1}{K_1}\), then \(C_2=\frac{K_2}{K_1} C_1\).
Substituting the given values, \(C_2=\frac{9}{3} C_1=3 C_1\).
Since the capacitor is isolated, the charge remains constant: \(Q_2=Q_0\).
Using the relationship \(Q=C V\), we have \(V=\frac{Q}{C}\).
The new voltage \(V_2\) is given by \(V_2=\frac{Q_2}{C_2}\).
Substituting \(Q_2=Q_0\) and \(C_2=3 C_1\), we get \(V_2=\frac{Q_0}{3 C_1}\).
Since \(V_0=\frac{Q_0}{C_1}\), then \(V_2=\frac{V_0}{3}\).
The electric field is given by \(E=\frac{V}{d}\).
The new electric field \(E_2\) is given by \(E_2=\frac{V_2}{d}\).
Substituting \(V_2=\frac{V_0}{3}\), we get \(E_2=\frac{V_0}{3 d}\).
Since \(E_0=\frac{V_0}{d}\), then \(E_2=\frac{E_0}{3}\).
The new values of charge, voltage, and electric field are \(Q_0, \frac{V_0}{3}\), and \(\frac{E_0}{3}\) respectively.

Hint

(d) Here capacitors of capacity \(4 \mu \mathrm{~F}\) each are in parallel, their equivalent capacity is \(8 \mu \mathrm{~F}\). Now, there is a combination of three capacitors in series of capacity \(20 \mu \mathrm{~F}, 8 \mu \mathrm{~F}\) and \(12 \mu \mathrm{~F}\) so, their resultant capacity,
\(
\begin{array}{ll}
& \frac{1}{C}=\frac{1}{20}+\frac{1}{8}+\frac{1}{12} \\
\therefore & C=\frac{120}{31} \mu \mathrm{~F}
\end{array}
\)
Total charge, \(Q=C V=\frac{120}{31} \times 300=1161 \mu \mathrm{C}\)
\(\therefore \quad\) Charge through \(4 \mu \mathrm{~F}\) capacitor \(=\frac{1161}{2}=580 \mu \mathrm{C}\)
and potential difference across \(4 \mu \mathrm{~F}\) condenser,
\(
V=\frac{q}{C}=\frac{580}{4}=145 \mathrm{~V}
\)

Hint

(c) As capacitors \(C_1\) and \(C_2\) are in series, then there sho equal charge on them, i.e.
\(
\begin{array}{rlrl}
& & \text { charge on } C_1 & =\text { charge on } C_2 \\
\therefore & C_1\left(V_A-V_D\right) & =C_2\left(V_D-V_B\right)
\end{array}
\)
\(
\begin{aligned}
C_1\left(V_1-V_D\right) & =C_2\left(V_D-V_2\right) \\
C_1 V_1-C_1 V_D & =C_2 V_D-C_2 V_2 \\
V_D\left(C_1+C_2\right) & =C_1 V_1+C_2 V_2
\end{aligned}
\)
\(
\begin{aligned}
&\therefore \quad \text { The potential difference of point } D \text {, }\\
&V_D=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}
\end{aligned}
\)

Hint

(a) Equivalent capacitance of circuit,

\(
C_{\mathrm{eq}}=\frac{6}{3+2+1}=1 \mu \mathrm{~F}
\)
Total charge, \(Q=1 \times 24=24 \mu \mathrm{C}\)
Now, potential difference across \(6 \mu \mathrm{~F}\) capacitor \(=\frac{24}{6}=4 \mathrm{~V}\)

Hint

(c) Given, circuit can be reduced as,

Let potential at \(P\) be \(V_P\) and potential at \(B\) be \(V_B\).
As capacitors \(3 \mu \mathrm{~F}\) and \(6 \mu \mathrm{~F}\) are in series, they have same charge.
\(
\begin{aligned}
& \therefore \quad \text { Charge on } 3 \mu \mathrm{~F}=\text { Charge on } 6 \mu \mathrm{~F} \\
& \therefore \quad C_1 V_1=C_2 V_2
\end{aligned}
\)
\(
3\left(1200-V_P\right)=6\left(V_P-V_B\right)
\)
\(
\begin{aligned}
&\text { As } B \text { point is attached to earth. }\\
&\begin{array}{lr}
\text { So, } & V_B=0 \\
\therefore & 1200-V_P=2 V_P \\
& V_P=400 \mathrm{~V}
\end{array}
\end{aligned}
\)

Hint

(b) The circuit can be redrawn as,

Here, \(4 \mu \mathrm{~F}\) and \(6 \mu \mathrm{~F}\) are in series. So, charge is same on both.
Now, equivalent capacity between \(A\) and \(B\),
\(
C_{A B}=\frac{6 \times 4}{6+4}=2.4 \mu \mathrm{~F}
\)
So, charge on \(4 \mu \mathrm{~F}\) capacitor,
\(
Q=C_{A B} \times 10=24 \times 10=24 \mu \mathrm{C}
\)

Hint

(d) Equivalent capacitance between \(A\) and \(B\),

\(
C_{A B}=C_1+\frac{C_1}{2}+C_1=\frac{5}{2} C_1
\)
As charge, \(Q=C V\)
So,
\(
\begin{aligned}
1.5 \times 10^{-6} & =\frac{5}{2} C_1 \times 6 \Rightarrow C_1=\frac{15}{15} \times 10^{-6} \\
& =0.1 \times 10^{-6} \mathrm{~F}=0.1 \mu \mathrm{~F}
\end{aligned}
\)

Hint

(c) Potential, \(V=\frac{q_{\text {net }}}{C_{\text {net }}}\)
\(
\Rightarrow \quad 40=\frac{(2)(200)}{2+C} \Rightarrow 80+40 C=400
\)
\(\therefore\) Capacity of second condenser, \(C=8 \mu \mathrm{~F}\)

Explanation:
The capacity of the first condenser is \(C_1=2 \mu \mathrm{~F}\).
The initial voltage across the first condenser is \(V_1=200 \mathrm{~V}\).
The final voltage across both condensers is \(V_f=40 \mathrm{~V}\).
When capacitors are connected in parallel, the voltage across them is the same.
The total charge in an isolated system remains constant.
The charge on a capacitor is given by \(Q=C V\).
The initial charge on the first condenser is given by:
\(Q_1=C_1 V_1\)
\(Q_1=(2 \mu \mathrm{~F})(200 \mathrm{~V})\)
\(Q_1=400 \mu \mathrm{C}\)
Since the battery is removed, the total charge remains constant.
The total charge is distributed between the two condensers:
\(Q_T=Q_1=400 \mu \mathrm{C}\)
The charge on the first condenser after connection is:
\(Q_1^{\prime}=C_1 V_f\)
\(Q_1^{\prime}=(2 \mu \mathrm{~F})(40 \mathrm{~V})\)
\(Q_1^{\prime}=80 \mu \mathrm{C}\)
The charge on the second condenser is the difference between the total charge and the charge on the first condenser:
\(Q_2=Q_T-Q_1^{\prime}\)
\(Q_2=400 \mu \mathrm{C}-80 \mu \mathrm{C}\)
\(Q_2=320 \mu \mathrm{C}\)
The capacity of the second condenser is:
\(C_2=\frac{Q_2}{V_f}\)
\(C_2=\frac{320 \mu \mathrm{C}}{40 \mathrm{~V}}\)
\(C_2=8 \mu \mathrm{~F}\)
The capacity of the second condenser is \(8 \mu \mathrm{~F}\).

Hint

\(
\text { (d) We have, } \quad q_i=q_f
\)
\(
\begin{aligned}
C_1 V_1+C_2 V_2 & =\left(C_1+C_2\right) V \\
(2)(4)+(C) 5 & =(2+C) 4.6 \\
0.4 C & =1.2 \\
C & =3 \text { units }
\end{aligned}
\)

Hint

(b) PD across \(6 \mu \mathrm{~F}\) capacitor,
\(
V_2=\left(\frac{3}{3+6}\right) \times 1200=\frac{3}{9} \times 1200=400 \mathrm{~V}
\)
Charge, \(q=4 \times 400=1600 \mu \mathrm{C}\)

Hint

(b) In parallel, potential difference is same and in series, it distributes in inverse ratio of capacity.

Potential difference,
\(
V=6\left(\frac{3}{3+7}\right)=1.8 \mathrm{~V}
\)
\(\therefore\) Charge stored in \(5 \mu \mathrm{~F}\) capacitor will be
\(
q=C V=(5)(1.8)=9 \mu \mathrm{C}
\)

Hint

(d) In series, potential difference distributes in inverse ratio of capacities. Hence,

\(
\begin{aligned}
V_1: V_2: V_3 & =\frac{1}{6}: \frac{1}{9}: \frac{1}{1.5}=1.5: 1: 6 \\
V_2 & =(3000)\left(\frac{1}{1+1.5+6}\right)=352.9 \mathrm{~V} \\
& \simeq 350 \mathrm{~V}
\end{aligned}
\)

Hint

(a) In the given figure capacitors \(3 \mu \mathrm{~F}, 3 \mu \mathrm{~F}\) and \(3 \mu \mathrm{~F}\) (between \(R\) and \(S\) ) are in series. If \(Q\) is the charge on each of these capacitors, then
\(
\begin{aligned}
&\begin{aligned}
& 30=\frac{Q}{3}+\frac{Q}{3}+\frac{Q}{3} \\
& Q=30 \mu \mathrm{C}
\end{aligned}\\
&\text { Potential difference between } R \text { and } S\\
&V=\frac{Q}{C}=\frac{30}{3}
\end{aligned}
\)
\(
V=10 \mathrm{~V}
\)

Hint

(c) The given circuit can be redrawn as follows

\(
\begin{aligned}
\left(V_A-V_B\right) & =\left(\frac{15}{5+15}\right) \times 2000 \\
\left(V_A-V_B\right) & =1500 \mathrm{~V} \\
2000-V_B & =1500 \mathrm{~V} \\
V_B & =500 \mathrm{~V}
\end{aligned}
\)

Hint

\(
\text { (c) Work done }=\frac{1}{2}\left(\frac{4 C}{3}\right) V^2=\frac{2 C V^2}{3}
\)

Hint

(c) Common potential, \(V=\frac{6 \times 20+3 \times 0}{(6+3)}=\frac{120}{9} \mathrm{~V}\) So, charge on \(3 \mu \mathrm{~F}\) capacitor (by closing \(S_2\) )
\(
Q_2=3 \times 10^{-6} \times \frac{120}{9}=40 \mu \mathrm{C}
\)

Hint

\(
\begin{aligned}
&\text { (b) The energy stored when the plates are fully charged, }\\
&U=\frac{1}{2}(2 C) V^2=C V^2=\left(\frac{\varepsilon_0 A}{d}\right)\left(V^2\right)
\end{aligned}
\)

Hint

(c) At the give positions all the charges are in equilibrium. But when they displaced slightly from their given position, they do not return back. So, they are in unstable equilibrium position.

Hint

\(
\begin{aligned}
&\text { (d) We have, } V \propto \frac{q}{R}\\
&\left(\because V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R}\right)
\end{aligned}
\)
\(
\begin{aligned}
&\begin{array}{ll}
\because & \sigma=\frac{q}{4 \pi R^2} \\
\therefore & q \propto \sigma R^2 \text { or } V \propto \sigma R
\end{array}\\
&\text { Potential is same, i.e. } V_1=V_2\\
&\begin{aligned}
& \therefore \quad \quad \sigma_1 R_1=\sigma_2 R_2 \\
& \Rightarrow \text { Ratio of } \frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}=\frac{5}{4}
\end{aligned}
\end{aligned}
\)

Hint

(c) \(V=\frac{q}{4 \pi \varepsilon_0 r}-\frac{q}{4 \pi \varepsilon_0 3 r}=\frac{q}{6 \pi \varepsilon_0 r}\)
Electric field intensity at a distance \(3 r\) is given by
Thus,
\(
\begin{aligned}
& E=\frac{q}{4 \pi \varepsilon_0(3 r)^2}=\frac{q}{4 \pi \varepsilon_0 \cdot 9 r^2} \\
& \frac{E}{V}=\frac{q\left(6 \pi \varepsilon_0 r\right)}{4 \pi \varepsilon_0 9 r^2(q)}=\frac{1}{6 r} \Rightarrow E=\frac{V}{6 r}
\end{aligned}
\)

Hint

\(
\text { (d) Area }=\frac{1}{2} Q V=\text { energy stored in the capacitor. }
\)

Hint

(c) Charge, \(q=n C V\)
\(
1=n \times 1 \times 10^{-6} \times 110
\)
Number of capacitors, \(n=\frac{1}{110 \times 10^{-6}}=\frac{100000}{11} \approx 9090\)

Hint

Hint

(a) By using, \(\frac{1}{2} m\left(v_1^2-v_2^2\right)=Q V\)
\(
\frac{1}{2} \times 10^{-3}\left[v_1^2-(0.2)^2\right]=10^{-8}(600-0)
\)
Velocity of the ball at the point \(A, v_1=22.8 \mathrm{cms}^{-1}\)

Hint

(c) Electric force, \(q E=m a\)
\(
\begin{array}{ll}
\therefore & a=\frac{q E}{m} \\
\Rightarrow & a=\frac{1.6 \times 10^{-19} \times 1 \times 10^3}{9 \times 10^{-31}}=\frac{1.6 \times 10^{15}}{9} \mathrm{~ms}^{-2}
\end{array}
\)
Initially speed, \(u=5 \times 10^6 \mathrm{~ms}^{-1}\) and final speed, \(v=0\)
\(\therefore\) From \(\quad v^2=u^2-2 a s\)
\(\Rightarrow\) Distance travelled by electron, \(s=\frac{u^2}{2 a}\)
\(
s=\frac{\left(5 \times 10^6\right)^2 \times 9}{2 \times 1.6 \times 10^{15}}=7 \mathrm{~cm}
\)

Hint

(a) Potential gradient is related with electric field according to the following relation, \(E=\frac{-d V}{d r}\).
\(
\begin{aligned}
\mathbf{E} & =-\frac{\partial V_{\mathbf{r}}}{\partial r}=\left[-\frac{\partial V}{\partial x} \hat{\mathbf{i}}-\frac{\partial V}{\partial y} \hat{\mathbf{j}}-\frac{\partial V}{\partial z} \hat{\mathbf{k}}\right] \\
& =\left[\hat{\mathbf{i}}\left(2 x y+z^3\right)+\hat{\mathbf{j}} x^2+\hat{\mathbf{k}} 3 x z^2\right]
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Potential energy of the system, }\\
&\frac{-k q Q}{x}-\frac{k Q q}{x}+\frac{k q^2}{2 x}=0
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow \frac{-4 k q Q+k q^2}{2 x}=0 \\
\Rightarrow k q^2=4 k Q q \\
\text { Ratio, } \frac{q}{Q}=4 &
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (c) Electrostatic potential energy, } U=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r}\\
&\begin{array}{ll}
\text { Here, } & q_1=q_2=1.6 \times 10^{-19} \mathrm{C} \\
\text { and } & r=9 \times 10^{-15} \mathrm{~m} \\
\therefore & U=\frac{9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{9 \times 10^{-15}} \\
& =2.56 \times 10^{-14} \mathrm{~J}
\end{array}
\end{aligned}
\)