Past NEET Entrance Papers

Hint

(b) The capacitance of a capacitor depends on the geometry of the capacitor, the distance between the plates, and the dielectric material between the plates.

The equation for capacitance ( \(C\) ) in terms of \(A\) and \(d\) is \(C=\varepsilon A / d\), where:
\(\boldsymbol{C}\) : The capacitance of the capacitor
\(\varepsilon\) : The permittivity of the dielectric material between the plates
A: The area of the parallel plates
\(d\) : The distance between the two conductive plates

Hint

(a) The electrostatic energy stored in a capacitor can be calculated using the formula:
\(
U=\frac{1}{2} C V^2
\)
Where:
\(U\) is the stored energy in joules (J)
\(C\) is the capacitance in farads (F)
\(V\) is the voltage in volts \((\mathrm{V})\)
Given:
\(
\begin{aligned}
& C=12 \mathrm{pF}\left(\text { which is } 12 \times 10^{-12} \mathrm{~F}\right) \\
& V=50 \mathrm{~V}
\end{aligned}
\)
Substitute the values into the formula:
\(
U=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2
\)
Calculate the values inside the parentheses first:
\(
50^2=2500
\)
Now substitute this back into the equation:
\(
U=\frac{1}{2} \times 12 \times 10^{-12} \times 2500
\)
Perform the multiplication:
\(
U=\frac{1}{2} \times 30000 \times 10^{-12}
\)
Which simplifies to:
\(
U=15000 \times 10^{-12}
\)
Convert this to nanojoules ( nJ ) by recognizing that \(1 \mathrm{~nJ}=10^{-9} \mathrm{~J}\) :
\(
U=15 \mathrm{~nJ}
\)

Hint

(c)  At steady state, capacitor will be completely charged and will not allow current to pass through it (open circuit).  The circuit simplifies to as shown below:

\(
i=\frac{V}{R} \Rightarrow i=\frac{10}{2+3}=2 \mathrm{~A}
\)

Hint

(d) Given circuit is balanced Wheatstone bridge, the equivalent circuit is shown below:

\(
\begin{aligned}
C_{A B} & =1+1 \\
& =2 \mu F
\end{aligned}
\)

Hint

(b) Given \(V^{\prime}=V=\) Constant
(i)
\(
\begin{aligned}
& C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d} \\
& d^{\prime}<d \\
& C^{\prime}>C
\end{aligned}
\)
Hence, final capacitance greater than initial capacitance,
(ii)
\(
\begin{aligned}
U^{\prime} & =\frac{1}{2} C^{\prime} V^2 \\
U & =\frac{1}{2} C V^2 \\
U^{\prime} & >U
\end{aligned}
\)
Hence final energy is greater than initial energy
(iii)
\(
\begin{aligned}
& \frac{Q^{\prime}}{V^{\prime}}=C^{\prime} \text { and } \frac{Q}{V}=C \\
& \frac{Q^{\prime}}{V^{\prime}} \neq \frac{Q}{V}
\end{aligned}
\)
(iv) Product of charge and voltage
\(
\begin{aligned}
& X^{\prime}=Q^{\prime} V=C^{\prime} V^2 \\
& X=Q V=C V^2 \\
& X^{\prime}>X
\end{aligned}
\)

Hint

(c) Substitute values into \(V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}\) :
\(
V= \pm \frac{2\left(4 \times 10^{-6}\right)}{4 \pi\left(9 \times 10^9\right)(2)^2}= \pm 9 \times 10^3 V
\)
Assertion A is true, but Reason R fails to consider directionality.

Hint

(c) A charged spherical shell has the same potential at all points inside the shell and on its surface.
This property arises because the electric field inside the shell is zero.
The potential \(V\) at any point inside the shell (including the center \(C\)) and on the surface \(P\) is given by:
\(
V=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{R}
\)
where
\(
\begin{aligned}
& q=1 \mu C=1 \times 10^{-6} C \\
& R=3 \mathrm{~cm}=0.03 \mathrm{~m}, \text { and } \\
& \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text { SIunits }
\end{aligned}
\)
Since the potential \(V\) is the same at both points \(C\) and \(P\),
the potential difference \(\Delta V=V_P-V_C=0\).
Thus, the potential difference between \(C\) and \(P\) is Zero.

Hint

(d) Step 1: Find the capacitance of the capacitor without the dielectric.
\(
C_0=\frac{\varepsilon_0 A}{d}
\)
Step 2: Find the capacitance of the capacitor with the dielectric.
The dielectric fills \(\frac{3}{4}\) of the space between the plates. The remaining \(\frac{1}{4}\) is filled with air.
\(
\begin{aligned}
& C=\frac{\varepsilon_0 A}{\frac{d}{4}}+\frac{3 \varepsilon_0 A}{\frac{3 d}{4}} \\
& C=\frac{4 \varepsilon_0 A}{d}+\frac{4 \varepsilon_0 A}{d} \\
& C=\frac{8 \varepsilon_0 A}{d} \\
& C=8 C_0
\end{aligned}
\)
Step 3: Use the potential difference equation to find the ratio of potential differences.
\(
\begin{aligned}
& V=\frac{Q}{C} \\
& \frac{V_0}{V}=\frac{\frac{Q}{C_0}}{\frac{Q}{C}} \\
& \frac{V_0}{V}=\frac{C}{C_0} \\
& \frac{V_0}{V}=\frac{8 C_0}{C_0} \\
& \frac{V_0}{V}=8 \\
& \frac{V_0}{V}=2: 1
\end{aligned}
\)

Hint

(d) Step 1: Convert the distance from centimeters to meters.
\(
r=9 \mathrm{~cm}=0.09 \mathrm{~m}
\)
Step 2: Substitute the given values into the formula for electric potential.
\(
\begin{aligned}
& V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\
& V=9 \cdot 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2} \cdot \frac{4 \cdot 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}
\end{aligned}
\)
Step 3: Simplify the expression.
\(
\begin{aligned}
& V=\frac{36 \cdot 10^2}{0.09} \mathrm{~V} \\
& V=4 \cdot 10^4 \mathrm{~V}
\end{aligned}
\)

Hint

(d) \(\mathrm{C}_{\mathrm{AB}}=\frac{3 \times 6}{3+6}=2 \mu \mathrm{~F}\)

Hint

(d) Three are in series and combination in parallel with other \(\frac{15}{3}+15=20 \mu \mathrm{~F}\)

Hint

(b)

\(
\begin{aligned}
& \mathrm{Q}=\mathrm{CV} \\
& \begin{aligned}
\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{C} \cdot \frac{\mathrm{dV}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}} & =\frac{\mathrm{I}}{\mathrm{C}}=\frac{2 \times 10^{-3}}{4 \times 10^{-6}} \\
& =\frac{10^3}{2}=500 \frac{\mathrm{~V}}{\mathrm{~s}}
\end{aligned}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\begin{aligned}
& v=\frac{k q}{2 \times 10^{-2}}-\frac{k q}{8 \times 10^{-2}} \\
& =k q\left[\frac{3}{8}\right] \times 10^{2}
\end{aligned}\\
&\text { So, the correct option is (b): }\left(\frac{3}{8}\right) q k
\end{aligned}
\)

Hint

(c) The electric potential \(V\) at the surface of a charged conducting sphere is given by the formula:
\(
V=\frac{k Q}{R}
\)
where \(k\) is Coulomb’s constant, \(Q\) is the charge on the sphere, and \(R\) is the radius of the sphere.
\(
Q=\frac{V R}{k}
\)
For a conducting sphere, the electric field \(E\) at a distance \(r\) from the center (where \(r>R\) ) can be calculated using the formula for the electric field due to a point charge:
\(
E=\frac{k Q}{r^2}
\)
Now substitute the expression for \(Q\) into the electric field formula:
\(
E=\frac{k\left(\frac{V R}{k}\right)}{r^2}
\)
Simplifying this gives:
\(
E=\frac{V R}{r^2}
\)

Hint

(c) The potential of the given conduction sphere is written as;
\(
\begin{aligned}
& \mathrm{V}=\frac{1}{4 \pi \epsilon_o} \frac{Q}{R} \\
& \Rightarrow \mathrm{~V}=\mathrm{k} \frac{Q}{R}
\end{aligned}
\)
Here V is the potential, Q is the charge, and k is the constant of proportionality.
CALCULATION:
According to the potential of the condition sphere we have;
\(
\mathrm{V}=\mathrm{k} \frac{Q}{R}
\)
Where R is inversely proportional to the applied potential.
In this question, we have two hollow conducting spheres where \(R_1 \gg R_2\). The larger the radius the smaller the potential for the given conducting spheres.
So, \(R_2\) is having large potential than \(R_1\).

Hint

(d) Concept:
We have considered an equipotential surface (a surface on which potential is constant everywhere).
\(\bar{n}=\) normal component through the surface
\(\vec{E}=\) electric field through a surface
\(\theta=\) angle between electric field and the surface of the sheet

\(
\text { here we know that potential difference }\Delta V=-\int \vec{E} \cdot d \vec{r}
\)

Here as we know that this is an equipotential surface then \(\Delta V=0\)
or, \(\Delta V=-\int \vec{E} \cdot d \vec{r}\)
from here we get \(E_{p e r} . d r=0\)
which says that the scalar product of the normal component of the electric field with distance is zero.
so that we can consider the normal component of the Electric field as \(=\mathrm{E} \cos \theta\) now, \(\mathrm{E} \cos \theta \cdot \mathrm{dr}=0\)
i.e. \(\mathbf{E}\) and \(\mathbf{d r}\) both are non-zero values
\(
\therefore \cos \theta=0 \Rightarrow \boldsymbol{\theta}=\mathbf{9 0}^{\circ}
\)
Now \(\theta=90^{\circ}\) means that the angle between electric lines of forces or electric field and the equipotential surface is \(90^{\circ}\).

Hint

(d) Concept:
Energy stored in capacitor \(=\frac{1}{2} C V^2=\frac{1}{2} Q V \dots(1)\)
Electrostatic energy stored in capacitor \(=\frac{1}{2}\left(C_1+C_2\right) V^2 \dots(2)\)
The common potential is \(V_c=\frac{C_1 V_1+C_2 V_2}{C_1+C_2} \dots(3)\)

Given: Voltage of battery \(=100 \mathrm{~V}\), Capacitance of capacitor \(=\mathrm{C}=900 \mathrm{pF}\)
Initially energy stored in capacitor \(=\frac{1}{2} C V^2=\frac{1}{2} Q V=(1 / 2) \times 900 \times 10^{-12} \times(100)^2=\) \(4.5 \times 10^{-6} \mathrm{~J}\)

When the capacitor is disconnected and is connected to another capacitor of 900 pF which can be seen as:

Charge on the system becomes \(Q^{\prime}=2 Q\)
From equation (3) we get:
The common potential is \(V_c=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\)
\(
\Rightarrow V_c=\frac{C \times 100+C \times 0}{C+C}=50 \mathrm{v}
\)
The electrostatic energy stored in capacitor \(=\frac{1}{2}\left(C_1+C_2\right) V^2=\) \(\frac{1}{2} 900 \times 10^{-12} \times 2 \times(50)^2=2.25 \times 10^{-6} \mathrm{~J}\)

Hint

(d) 1. Understanding the Properties of a Hollow Conducting Sphere:
A hollow conducting sphere has the property that the electric field inside the sphere is zero. This means that there are no electric field lines within the hollow region of the sphere.
2. Electric Potential Inside the Sphere:
Since the electric field inside the hollow conducting sphere is zero, the electric potential throughout the interior of the sphere is constant. This constant potential is equal to the potential at the surface of the sphere.
3. Calculating the Surface Potential:
The electric potential \(V\) at the surface of a charged sphere is given by the formula:
\(
V=k \frac{Q}{R}
\)
where \(k=\frac{1}{4 \pi \epsilon_0}\).
4. Substituting the Values:
Therefore, the potential at the surface of the sphere is:
\(
V=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R}
\)
5. Conclusion:
Since the potential inside the sphere is uniform and equal to the surface potential, the electric potential at a distance \(r=\frac{R}{3}\) from the center of the sphere is also:
\(
V=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R}
\)

Hint

(b) The equivalent capacitance,
\(
\begin{aligned}
& C=C_1+C_2+C_3 \\
& =(0.3+0.3+0.3)=0.9 \mu F
\end{aligned}
\)
This combination is connected with another capacitance \(0.1 \mu F\) in series, then
\(
\begin{aligned}
\frac{1}{C } & =\frac{1}{0.9}+\frac{1}{0.1}=\frac{1+9}{0.9} \\
\frac{1}{C } & =\frac{10}{0.9}
\end{aligned}
\)
Resultant capacitance, \(C=\frac{0.9}{10}=0.09 \mu F\)

Hint

(a) Let the closest distance of approach be \(r^{\prime}\). From the principle of conservation of energy we have,
kinetic energy = electrostatic potential energy.
In the first case,
\(\frac{1}{2} m v^2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q Q}{r} \dots(1)\).
In the second case,
\(\frac{1}{2} m(2 v)^2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q Q}{r^{\prime}} \dots(2)\).
Dividing equation (1) by equation (2) we get,
\(
\frac{1}{4}=\frac{r^{\prime}}{r} \text { or }, r^{\prime}=\frac{r}{4}
\)

Hint

(d) For a capacitor of area A and distance between the plates as \(d\), the capacitance (C) is
\(
C=\frac{A \epsilon_0}{d}
\)
On doubling the distance and reducing area to half
\(
\begin{aligned}
& C_1=\frac{\frac{A}{2} \epsilon_0}{2 d} \\
& \Rightarrow C_1=\frac{A \epsilon_0}{4 d} \\
& \Rightarrow C_1=\frac{C}{4}
\end{aligned}
\)

Hint

(c) Here, \(\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{3}\) (Series combination)
While, \(C_1+C_2=16\) (Parallel combination)
Solving the above equations we get
\(
C_1=12 \mu F \text { and } C_2=4 \mu F
\)

Hint

(a) 

\(
\begin{aligned}
& \text { Work done }=q_0 \cdot\left(V_0-V_{\infty}\right) \\
& =\left\{\frac{3 k q}{d}+\left(\frac{-3 k q}{d}\right)\right\} q_0 \\
& =\text { Zero }
\end{aligned}
\)

Hint

(d)

\(
C_{A B}=2 C
\)

Hint

(d)

\(
\begin{aligned}
& \frac{q_1}{q_2}=\frac{C_1}{C_2}=\frac{R_1}{R_2} \\
& \text { and } \frac{q_1}{q_2}=\frac{4 \pi R_1^2 \sigma_1}{4 \pi R_2^2 \sigma_2} \\
& \frac{R_1}{R_2}=\frac{4 \pi R_1^2 \sigma_1}{4 \pi R_2^2 \sigma_2} \\
& \therefore \frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}
\end{aligned}
\)

Hint

(b) If each drop has a charge ‘ \(q\) ‘ and radius ‘ \(r\) ‘.
Then from conservation of charge, charge on the big drop is \(n q=27 q(n=27)\)
from conservation of volume \(\frac{4}{3} \pi r^3 n=\frac{4}{3} \pi R^3\)
\(
R=n^{1 / 3} r
\)
Now potential of the small drop \(V=\frac{q}{4 \pi \epsilon_0 r}=220 \mathrm{~V}\)
Potential of the big drop,
\(
\begin{aligned}
& V=\frac{n q}{4 \pi \epsilon_0 R}=\frac{n q}{4 \pi \epsilon_0 n^{1 / 3} r}=n^{2 / 3} \frac{q}{4 \pi \epsilon_0 r} \\
& V=(27)^{2 / 3} \times 220 \mathrm{~V} \\
& =9 \times 220=1980 \mathrm{~V}
\end{aligned}
\)

Hint

(d) The energy stored by a capacitor
\(
U=\frac{1}{2} C V^2 \dots(i)
\)
\(V\) is the p.d. between two plates of the capacitor potential difference \(\quad V=E . d\).
The capacitance of the parallel plate capacitor
\(
C=\frac{A \varepsilon_0}{d}
\)
Substituting the value of \(C\) in equation (i)
\(
U=\frac{1}{2} \frac{A \varepsilon_0}{d}(E d)^2=\frac{1}{2} A \varepsilon_0 E^2 d
\)

Hint

(d) Since, electric potential is constant throughout the volume, hence electric field,
\(
E=-\frac{d V}{d r}=0
\)

Hint

(a) Given,
Dipole moment of short electric dipole, \(p=16 \times 10^{-9} \mathrm{Cm}\).
Distance from centre of dipole, \(r=0.6 \mathrm{~m}\)
Electric potential, \(V=\frac{k p \cos \theta}{r^2}\)
\(
\Rightarrow V=\frac{9 \times 10^9 \times 16 \times 10^{-9} \times \cos 60^{\circ}}{0.36}=200 \mathrm{~V}
\)

Hint

(b) Capacitance of a parallel plate capacitor with air is
\(
C=\frac{\varepsilon_0 A}{d} \dots(i)
\)
Here, \(A=\) area of plates of capacitor, \(d=\) distance between the plates
Capacitance of a same parallel plate capacitor with introduction of dielectric medium of dielectric constant \(K\) is
\(
C^{\prime}=\frac{K \varepsilon_0 A}{d} \dots(ii)
\)
Dividing (ii) by (i)
\(
\begin{aligned}
& \Rightarrow \frac{C^{\prime}}{C}=K \Rightarrow \frac{30}{6}=K \Rightarrow K=5 \\
& \Rightarrow K=\frac{\varepsilon}{\varepsilon_0} \\
& \Rightarrow \varepsilon=K \varepsilon_0=5 \times 8.85 \times 10^{-12} \\
& \quad=0.44 \times 10^{-10} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}
\end{aligned}
\)

Hint

(b)
\(
\text { For } r<R \quad ; \quad V=\frac{q}{4 \pi \varepsilon_0 R}=\text { costant }
\)
\(
\text { for } r \geq R ; \quad V=\frac{q}{4 \pi \varepsilon_0 r}=V \propto \frac{1}{r}
\)

Hint

(b) Capacitance of parallel plate capacitor when medium is air
\(
\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \dots(i)
\)
According to second condition,
\(
\mathrm{A}^{\prime}=\mathrm{A}, \mathrm{t}=\mathrm{d} / 2, \mathrm{~K}=4
\)
\(\therefore\) Capacitance, \(C=\frac{\varepsilon_0 A}{(d-t)+\frac{t}{\kappa}}=\frac{\varepsilon_0 A}{\left(d-\frac{d}{2}\right)+\frac{d / 2}{4}}\)
\(
=\frac{\varepsilon_0 \mathrm{~A}}{\frac{\mathrm{~d}}{2}+\frac{\mathrm{d}}{8}}=\frac{8}{5} \cdot \frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}
\)
\(
\begin{array}{lc}
\therefore & \frac{\mathrm{C}}{\mathrm{C}_0}=\frac{\frac{8}{5} \cdot \frac{\varepsilon_0 \Lambda}{d}}{\frac{\frac{8}{0} \Lambda}{d}} \\
\Rightarrow & \frac{\mathrm{C}}{\mathrm{C}_0}=\frac{8}{5}
\end{array}
\)

Hint

(d) Charge \(\mathbf{Q}\) with a surface charge density \(\boldsymbol{\sigma}\) on a spherical body of radius \(\mathbf{R}\) is given as:
\(
\mathrm{Q}=\sigma \times 4 \pi \mathrm{R}^2 \dots(1)
\)
Potential of a sphere with charge Q and radius R is given as:
\(
\mathrm{V}=\frac{Q}{4 \pi \epsilon_0 R} \dots(2)
\)
Given:
Radius of 1st sphere \(=R\)
Radius of 2 nd sphere \(=2 \mathrm{R}\)
Surface charge density on both 1st and 2nd sphere \(=\sigma\)
Now, suppose initial charge on 1st sphere is \(\mathrm{Q}_1\) and that on 2nd sphere is \(\mathrm{Q}_2\).
Using equation (1) we have,
\(
\mathrm{Q}_1=\sigma \times 4 \pi \mathrm{R}^2
\)
\(\mathrm{Q}_2=\sigma \times 4 \pi(2 R)^2=4 \mathrm{Q}_1 \dots(3)\)
After separating shperes when brought in contact both have same potential.
Let us suppose final charges on two spheres are \(Q_1^{\prime} \& Q^{\prime}{ }_2\).
\(
\begin{aligned}
& \therefore \mathrm{V}_1=\mathrm{V}_2 \Rightarrow \frac{Q_1^{\prime}}{4 \pi \epsilon_0 R}=\frac{Q_2^{\prime}}{4 \pi \epsilon_0(2 R)} \quad \text { (using equation (2)) } \\
& \Rightarrow \mathrm{Q}_2^{\prime}=2 \mathrm{Q}_1^{\prime} \dots(4)
\end{aligned}
\)
\(\because\) Total charge on the sphere will be same before and after separation.
\(
\begin{aligned}
& \therefore Q_1+Q_2=Q_1^{\prime}+Q_2^{\prime} \\
& \Rightarrow Q_1+4 Q_1=Q_1^{\prime}+2 Q_1^{\prime} \\
& \Rightarrow 5 Q_1=3 Q_1^{\prime} \Rightarrow Q_1^{\prime}=\frac{5}{3} Q_1=\frac{5}{3} \times \sigma \times 4 \pi R^2
\end{aligned}
\)
or, \(\sigma_1=\frac{Q_1^{\prime}}{4 \pi \epsilon_0 R^2}=\frac{5}{3} \sigma\)
Simillarly, \(\boldsymbol{\sigma}_2=\frac{5}{6} \boldsymbol{\sigma}\)

Hint

\(
\begin{aligned}
&\text { (d) : As we know that, loss of electrostatic energy, }\\
&\begin{aligned}
E_{\text {loss }} & =\frac{1}{2} \frac{C_1 C_2}{\left(C_1+C_2\right)} V^2=\frac{1}{2} \times \frac{C^2}{2 C} V^2 \\
& =\frac{1}{2}\left(\frac{1}{2} C V^2\right)=\frac{1}{2} E \quad \quad\left[\because C_1=C_2=C\right] \\
\therefore \quad & \text { Percentage of loss of energy }=\frac{1}{2} E \\
& =\frac{1}{2} \times 100 \%=50 \% .
\end{aligned}
\end{aligned}
\)

Hint

(c) Capacitor-It is a two-terminal electrical device that stores energy as an electric charge.
The capacitance of the capacitor is the ratio of the charge per applied potential and it is written as;
\(
C=\frac{q}{V} \dots(1)
\)
Here, Q is the charge and V is the applied potential.
Given that: Capacitance of capacitor \(\mathrm{C}=20 \mu \mathrm{~F}\)
\(
C=20 \times 10^{-6} \mathrm{~F}
\)
and Rate of change of potential \(\left(\frac{d V}{d t}\right)=3 \mathrm{v} / \mathrm{s}\)
Now, on using the equation (1) we have;
\(
\mathrm{q}=\mathrm{CV} \dots(2)
\)
Differentiate the equation (2) with respect to time \(t\) we have;
\(
\frac{d q}{d t}=C \frac{d V}{d t} \dots(3)
\)
As we know the rate to change of charge per unit of time is equal to the current, therefore equation 3 ) is
\(
i_c=C \frac{d V}{d t}
\)
Now, on putting all the given values in equation (3) we have;
\(
\begin{aligned}
i_C & =20 \times 10^{-6} \times 3 \\
& =60 \times 10^{-6} \mathrm{~A} \\
& =60 \mu \mathrm{~A}
\end{aligned}
\)
As we know that displacement current is equal to capacitive current we have,
\(
i_d=i_c=60 \mu \mathrm{~A}
\)

Hint

(a) 1. Understand the Setup: We have a parallel plate capacitor with two plates. One plate has charge \(+Q\) and the other has charge \(-Q\). The area of each plate is \(A\).
2. Calculate the Electric Field: The electric field \(E\) between the plates of a parallel plate capacitor can be calculated using the formula:
\(
E=\frac{\sigma}{\epsilon_0}
\)
where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space. The surface charge density \(\sigma\) is given by:
\(
\sigma=\frac{Q}{A}
\)
Therefore, the electric field \(E\) becomes:
\(
E=\frac{Q}{A \epsilon_0}
\)
3. Calculate the Force on One Plate: The force \(F\) on one plate due to the electric field created by the other plate can be calculated using the formula:
\(
F=Q \cdot E
\)
Substituting the expression for \(E\) :
\(
F=Q \cdot\left(\frac{Q}{2 A \epsilon_0}\right)=\frac{Q^2}{2 A \epsilon_0}
\)
Here, we use \(\frac{Q}{2 A \epsilon_0}\) because the electric field due to one plate at the location of the other plate is half of the total electric field.
4. Independence from Distance:
Notably, this expression shows that the force is independent of the distance \(d\) between the plates.
Final Answer:
The electrostatic force between the metal plates of an isolated parallel plate capacitor is given by:
\(
F=\frac{Q^2}{2 A \epsilon_0}
\)

Hint

(a) As the regions are of equipotentials, so Work done \(\mathrm{W}=\mathrm{q} \Delta \mathrm{V}\).
Since \(\Delta V\) is the same in all the cases hence work done will also be the same in all the cases.

Hint

(a) When battery is replaced by another uncharged capacitor, As uncharged capacitor is connected parallel So, \(\quad C^{\prime}=2 \mathrm{C}\)

\(
\text { Common potential, } V^{\prime}=\frac{C V}{C+C}=\frac{V}{2}
\)
When the capacitor is charged by a battery of potential \(V\), then energy stored in the capacitor,
\(
\text { Initial Energy of system, } U_i=\frac{1}{2} \mathrm{CV}^2 \dots(i)
\)
\(
\begin{aligned}
&\text { Then the energy stored in the capacitor }\\
&U_f=\frac{1}{2}(2 C)\left(\frac{V}{2}\right)^2=\frac{1}{4} C V^2 \dots(ii)
\end{aligned}
\)
\(\therefore\) From eqns. (i) and (ii)
\(
U_f=\frac{U_i}{2}
\)
It means the total electrostatic energy of resulting system will decreases by a factor of 2.

Hint

(b) 

\(
\begin{aligned}
& \text { Here } \mathrm{E}=2 \times 10^5 \mathrm{~NC}^{-1}, l=2 \mathrm{~cm}, \tau=4 \mathrm{Nm} \\
& \text { Torque } \vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}=\mathrm{pE} \sin \theta \\
& \therefore 4=\mathrm{p} \times 2 \times 10^5 \times \sin 30^{\circ} \\
& \text { or } \mathrm{p}=4 \times 10^{-5} \mathrm{~Cm} \\
& \therefore \text { Charge } q=\frac{p}{l}=\frac{4 \times 10^{-5} \mathrm{~Cm}}{0.02 \mathrm{~m}}=2 \times 10^{-3} \mathrm{C}=2 \mathrm{~mC}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (c) } \text { Here, } C_1=\frac{2 \varepsilon_0 k_1 A}{3 d}, C_2=\frac{2 \varepsilon_0 k_2 A}{3 d} \\
& C_3=\frac{2 \varepsilon_0 k_3 A}{3 d}, C_4=\frac{2 \varepsilon_0 k_4 A}{d}
\end{aligned}
\)
\(
\text { Given system of } C_1, C_2, C_3 \text { and } C_4 \text { can be simplified as }
\)

\(
\begin{aligned}
& \therefore \quad \frac{1}{C_{A B}}=\frac{1}{C_1+C_2+C_3}+\frac{1}{C_4} \\
& \text { Suppose, } C_{A B}=\frac{k \varepsilon_0 A}{d} \\
& \frac{1}{k\left(\frac{\varepsilon_0 A}{d}\right)}=\frac{1}{\frac{2}{3} \frac{\varepsilon_0 A}{d}\left(k_1+k_2+k_3\right)}+\frac{1}{\frac{2 \varepsilon_0 A}{d} k_4} \\
& \Rightarrow \frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2 k_4} \therefore \frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}
\end{aligned}
\)

Hint

(d) When S and 1 are connected
The \(2 \mu \mathrm{~F}\) capacitor gets charged. The potential difference across its plates will be V . The potential energy stored in \(2 \mu \mathrm{~F}\) capacitor
\(
U_i=\frac{1}{2} C V^2=\frac{1}{2} \times 2 \times V^2=V^2
\)
When S and 2 are connected
The \(8 \mu \mathrm{~F}\) capacitor also gets charged. During this charging process current flows in the wire and some amount of energy is dissipated as heat. The energy loss is
\(
\begin{aligned}
& \quad \Delta U=\frac{1}{2} \frac{\mathrm{C}_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2 \\
& \text { Here, } C_1=2 \mu F, C_2=8 \mu F, V_1=V, V_2=0 \\
& \therefore \Delta U=\frac{1}{2} \times \frac{2 \times 8}{2+8}(V-0)^2=\frac{4}{5} V^2
\end{aligned}
\)
The percentage of the energy dissipated
\(
=\frac{\Delta U}{U_i} \times 100=\frac{\frac{4}{5} V^2}{V^2} \times 100=80 \%
\)

Hint

(a) Potential in a region is given by,
\(
V=6 x y-y+2 y z
\)
As we know the relation between electric potential and electric field is
\(
\begin{aligned}
& \overrightarrow{\mathrm{E}}=-\left(\frac{\partial V}{\partial x} \hat{\mathrm{i}}+\frac{\partial \mathrm{V}}{\partial y} \hat{\mathrm{j}}+\frac{\partial \mathrm{V}}{\partial z} \hat{\mathrm{k}}\right) \\
& \text { where } \mathrm{V}=6 x y-y+2 y z \\
& \overrightarrow{\mathrm{E}}=-[(6 y \hat{i}+(6 x-1+2 z) \hat{\mathrm{j}}+(2 y) \hat{\mathrm{k}}] \\
& \overrightarrow{\mathrm{E}}_{(1,1,0)}=-(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})
\end{aligned}
\)

Hint

(d) Force of attraction between the plates, \(\mathrm{F}=\mathrm{qE}\)
\(
\begin{aligned}
& =q \times \frac{\sigma}{2 \epsilon_0} \\
& =q \frac{q}{2 A \epsilon_0} \quad\left[\because E=\frac{6}{2 E_0}\right] \\
& =\frac{q^2}{2\left(\frac{\epsilon_0 A}{d}\right) \times d}=\frac{C^2 V^2}{2 C d}=\frac{C V^2}{2 d}
\end{aligned}
\)
Here, \(C=\frac{\epsilon_0 A}{d}, q=C V, A=\) area
Note: To find the force of attraction between the plates, we use the concept that work done in displacing the plates against the force is equal to the increase in energy of the capacitor.

Hint

(c) Capacitance of the capacitor, \(C=\frac{Q}{V}\)
After inserting the dielectric, new capacitance
\(
\mathrm{C}^{\prime}=\mathrm{K} \cdot \mathrm{C}
\)
New potential difference
\(
\begin{aligned}
& \mathrm{V}^{\prime}=\frac{\mathrm{V}}{\mathrm{~K}} \\
& \mathrm{U}_{\mathrm{i}}=\frac{1}{2} \mathrm{CV}^2=\frac{\mathrm{Q}^2}{2 \mathrm{C}} \quad(\because \mathrm{Q}=\mathrm{CV}) \\
& \mathrm{U}_{\mathrm{f}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}}=\frac{\mathrm{Q}^2}{2 \mathrm{KC}}=\frac{\mathrm{C}^2 \mathrm{~V}^2}{2 \mathrm{KC}}=\left(\frac{\mathrm{U}_{\mathrm{i}}}{\mathrm{~K}}\right) \\
& \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=\frac{1}{2} \mathrm{CV}^2\left\{\frac{1}{\mathrm{~K}}-1\right\}
\end{aligned}
\)
As the capacitor is isolated, so charge will remain conserved and p.d. between two plates of the capacitor
\(
\mathrm{V}^{\prime}=\frac{\mathrm{Q}}{\mathrm{KC}}=\frac{\mathrm{V}}{\mathrm{~K}}
\)

Hint

(c) Electric field, \(\mathrm{E} \propto \frac{1}{\mathrm{~K}}\)
As \(\mathrm{K}_1<\mathrm{K}_2\) so \(\mathrm{E}_1>\mathrm{E}_2\) Hence graph (c) correctly dipicts the variation of electric field \(E\) with distance \(d\).

Hint

(b) For the conducting sphere,
\(
\begin{aligned}
\text { Potential at the centre } & =\text { Potential on the sphere } \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}
\end{aligned}
\)
Electric field at the centre \(=0\)

Hint

\(
\begin{aligned}
& \text { (d) } \overrightarrow{\mathrm{E}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} \hat{\mathrm{i}}-\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \hat{\mathrm{j}}-\frac{\partial \mathrm{V}}{\partial \mathrm{z}} \hat{\mathrm{k}} \\
& \text { where } \mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{x}-8 \mathrm{xy}-8 \mathrm{y}+6 \mathrm{yz} \\
& =-[(6-8 \mathrm{y}) \hat{\mathrm{i}}+(-8 \mathrm{x}-8+6 \mathrm{z}) \hat{\mathrm{j}}+(6 \mathrm{y}) \hat{\mathrm{k}}] \\
& \text { At }(1,1,1), \overrightarrow{\mathrm{E}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\
& \Rightarrow \quad(\overrightarrow{\mathrm{E}})=\sqrt{(2)^2+(10)^2+(-6)^2}=\sqrt{140}=2 \sqrt{35} \\
& \therefore \quad \mathrm{~F}=\mathrm{q} \overrightarrow{\mathrm{E}}=2 \times 2 \sqrt{35}=4 \sqrt{35} N
\end{aligned}
\)

Hint

(d) In the direction of electric field, electric potential decreases.
\(
\therefore \quad V_B>V_C>V_A
\)

Hint

(b) Potential energy of dipole,
\(
U=-p E\left(\cos \theta_2-\cos \theta_1\right)
\)
Here, \(\theta_1=0^{\circ}, \theta_2=90^{\circ}\)
\(
\therefore \quad U=-p E\left(\cos 90^{\circ}-\cos 0^{\circ}\right)=-p E(0-1)=p E
\)

Hint

(a) The torque \(\tau\) experienced by an electric dipole in an electric field is given by the formula:
\(
\tau=\mathbf{p} \times \mathbf{E}
\)
Where \(\mathbf{p}\) is the dipole moment vector and \(\mathbf{E}\) is the electric field vector.
Step 2: Calculate the Magnitude of Torque
The magnitude of the torque can be expressed as:
\(
\tau=p E \sin \theta
\)
Here, \(p\) is the magnitude of the dipole moment, \(E\) is the magnitude of the electric field, and \(\theta\) is the angle between the dipole moment and the electric field.
Step 3: Understand the Potential Energy of the Dipole
The potential energy \(U\) of an electric dipole in an electric field is given by the formula:
\(
U=-\mathbf{p} \cdot \mathbf{E}
\)
This can also be expressed in terms of the angle \(\theta\) :
\(
U=-p E \cos \theta
\)
Step 4: Substitute the Values
Since we have assumed that the potential energy is zero when \(\theta=90^{\circ}\), we can directly use the formulas derived:
The torque is:
\(
\tau=p E \sin \theta
\)
The potential energy is:
\(
U=-p E \cos \theta
\)

Hint

(a) Let \(a\) be the side of the square \(A B C D\).

\(
\begin{aligned}
& \therefore \quad A C=B D=\sqrt{a^2+a^2}=a \sqrt{2} \\
& O A=O B=O C=O D=\frac{a \sqrt{2}}{2}=\frac{a}{\sqrt{2}}
\end{aligned}
\)
Potential at the centre \(O\) due to given charge configuration is
\(
V=\frac{1}{4 \pi \varepsilon_0}\left[\frac{(-Q)}{\left(\frac{a}{\sqrt{2}}\right)}+\frac{(-q)}{\left(\frac{a}{\sqrt{2}}\right)}+\frac{(2 q)}{\left(\frac{a}{\sqrt{2}}\right)}+\frac{2 Q}{\left(\frac{a}{\sqrt{2}}\right)}\right]=0 \text { (Given) }
\)
\(
\Rightarrow-Q-q+2 q+2 Q=0 \text { or } Q+q=0 \text { or } Q=-q
\)

Hint

(b) for equilibrium
\(
\frac{d U}{d r}=0 \quad \Rightarrow \quad \frac{-2 A}{r^3}+\frac{B}{r^2}=0 \quad r=\frac{2 A}{B}
\)
for stable equilibrium
\(\frac{d^2 U}{d r^2}\) should be positive for the value of \(r\).
Here \(\frac{d^2 U}{d r^2}=\frac{6 A}{r^4}-\frac{2 B}{r^3}\) is + ve value for
\(
r=\frac{2 A}{B}
\)
Note: After displacing a charged particle from its equilibrium position, if it return back then it is said to be in stable equilibrium. If \(\mathrm{U}\) is the potential energy then in case of stable equilibrium \(\frac{d^2 U}{d x^2}\) is positive \(i . e, \mathrm{U}\) is minimum.

Hint

(a) When the given metallic spheres are connected by a conducting wire, charge will flow till both the spheres acquire a common potential which is given by
\(
\begin{aligned}
V & =\frac{q_1+q_2}{C_1+C_2}=\frac{-1 \times 10^{-2}+5 \times 10^{-2}}{4 \pi \varepsilon_0 R_1+4 \pi \varepsilon_0 R_2} \\
& =\frac{4 \times 10^{-2}}{4 \pi \varepsilon_0\left(1 \times 10^{-2}+3 \times 10^{-2}\right)}=\frac{1}{4 \pi \varepsilon_0}
\end{aligned}
\)
\(\therefore \quad\) Final charge on the bigger sphere \(=\mathrm{C}_2 \mathrm{~V}\)
\(
=4 \pi \varepsilon_0 \times 3 \times 10^{-2} \times \frac{1}{4 \pi \varepsilon_0}=3 \times 10^{-2} \mathrm{C}
\)

Hint

(d) In series, \(\mathrm{C}_{\mathrm{eff}}=\frac{C_1}{n_1}\)
\(\therefore \quad\) Energy stored,
\(
\begin{aligned}
& E_S=\frac{1}{2} C_{e f f} V_S^2=\frac{1}{2} \frac{C_1}{n_1} 16 V^2 \\
& =8 V^2 \frac{C_1}{n_1} \\
& \text { In parallel, } \mathrm{C}_{\mathrm{eff}}=\mathrm{n}_2 \mathrm{C}_2
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \quad \text { Energy stored, } \mathrm{E}_{\mathrm{p}}=\frac{1}{2} \mathrm{n}_2 \mathrm{C}_2 \mathrm{~V}^2 \\
& \text { According to question } \mathrm{E}_{\mathrm{s}}=\mathrm{E}_{\mathrm{p}} \\
& \therefore \quad \frac{8 V^2 C_1}{n_1}=\frac{1}{2} n_2 C_2 V^2 \\
& \Rightarrow \quad C_2=\frac{16 C_1}{n_1 n_2}
\end{aligned}
\)

Hint

(b) The electric field between two charged parallel plates placed in air is
\(
E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0 A} \ldots(i)
\)
When the parallel plates are immersed in kerosene oil tank of dielectric constant \(K\), the electric field between the plates is
\(
\begin{aligned}
& E^{\prime}=\frac{\sigma}{K \varepsilon_0}=\frac{Q}{K \varepsilon_0 A}=\frac{E}{K} \text { (Using (i)) } \\
& \therefore \frac{E^{\prime}}{E}=\frac{1}{K}
\end{aligned}
\)
As \(K>1\), so \(E^{\prime}<E\)

Hint

(c) Three capacitors of capacitance \(C\) each are in series.
\(\therefore\) Total capacitance, \(C_{\text {total }}=\frac{C}{3}\)
The charge is the same, \(Q\), when capacitors are in series.
\(
V_{\text {total }}=\frac{Q}{C}=\frac{Q}{C / 3}=3 \mathrm{~V}
\)

Hint

(b)  As the capacitors are connected in parallel, therefore potential difference across both the condensers remains the same.
\(
\begin{aligned}
\therefore \quad Q_1 & =C V \\
Q_2 & =\frac{C}{2} V \\
\text { Also, } Q & =Q_1+Q_2 \\
=C V & +\frac{C}{2} V=\frac{3}{2} C V
\end{aligned}
\)
Work done in charging fully both the condensers is given by \(W=\frac{1}{2} Q V=\frac{1}{2} \times\left(\frac{3}{2} C V\right) V=\frac{3}{4} C V^2\).

Hint

(d) If we increase the distance between the plates its capacity decreases resulting in higher potential as we know \(Q=C V\). Since \(Q\) is constant (battery has been disconnected), on decreasing \(C, V\) will increase.

Hint

(b) \(C_1, C_2\) and \(C_3\) are in series
\(
\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C}
\)
or, \(\frac{1}{C^{\prime}}=\frac{6+3+2}{6 C}=\frac{11}{6 C}\) or, \(C^{\prime}=\frac{6 C}{11}\)
As the capacitors \(C_1, C_2\) and \(C_3\) are in series so the charge on each capacitor is \(Q^{\prime}=\frac{6}{11} C V\)
Also charge on capacitor \(C_4\) is \(Q=C_4 V=4 C V\)
\(
\therefore \text { Ratio }=\frac{Q^{\prime}}{Q}=\frac{6 C V}{11 \times 4 C V}=\frac{3}{22}
\)

Hint

(c)

To get equivalent capacitance \(6 \mu \mathrm{~F}\). Out of the \(4 \mu \mathrm{~F}\) capacitance, two are connected in series and third one is connected in parallel.
\(
C_{\mathrm{eq}}=\frac{4 \times 4}{4+4}+4=2+4=6 \mu \mathrm{~F}
\)

Hint

(b) Charge on first capacitor \(=q_1=C_1 V\)
Charge on second capacitor \(=q_2=0\)
When they are connected, in parallel the total charge
\(
q=q_1+q_2 \quad \therefore q=C_1 V
\)
and capacitance, \(C=C_1+C_2\)
Let \(V^{\prime}\) be the common potential difference across each capacitor, then \(q=C V^{\prime}\).
\(
\therefore \quad V^{\prime}=\frac{q}{C}=\frac{C_1}{C_1+C_2} V
\)

Hint

(a) Electric potential is constant (equal to \(\frac{k q}{R}\), where \(k=\frac{1}{4 \pi \epsilon_0}\) ) within or on the surface of conductor.

Hint

\(
\begin{aligned}
&\text { (a) Energy stored in capacitor }\\
&=\frac{1}{2} C V^2=\frac{1}{2} \frac{Q}{V} V^2 \quad(Q=C V)=\frac{1}{2} Q V
\end{aligned}
\)

Hint

(c) \(C=10 \mu \mathrm{~F} \quad d=8 \mathrm{~cm}\)
\(
\begin{aligned}
& C^{\prime}=? \quad d^{\prime}=4 \mathrm{~cm} \\
& C=\frac{A \epsilon_0}{d} \quad \rightarrow \quad C \propto \frac{1}{d}
\end{aligned}
\)
If d is halved then C will be doubled.
Hence, \(C^{\prime}=2 C=2 \times 10 \mu F=20 \mu F\)

Hint

\(
\begin{aligned}
&\text { (a) Energy stored per unit volume }\\
&=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2=\frac{1}{2} \varepsilon_0 \frac{V^2}{d^2}\left(\because E=\frac{V}{d}\right)
\end{aligned}
\)

Hint

(b) Let \(q\) be the charge on each capacitor.
\(\therefore\) Energy stored, \(U=\frac{1}{2} C V^2=\frac{1}{2} \frac{q^2}{C}\)
Now, when battery is disconnected and another capacitor of same capacity is connected in parallel to the first capacitor, then voltage across each capacitor, \(V=\frac{q}{2 C}\)
\(\therefore\) Energy stored \(=\frac{1}{2} C\left(\frac{q}{2 C}\right)^2=\frac{1}{4} \cdot \frac{1}{2} \frac{q^2}{C}=\frac{1}{4} U\)

Hint

(c) Step 1: Determine the initial capacitance
The capacitance \(C\) of a parallel plate capacitor is given by the formula:
\(
C=\frac{A \epsilon_0}{d}
\)
where:
\(\boldsymbol{A}\) is the area of the plates,
\(\epsilon_0\) is the permittivity of free space,
\(d\) is the separation between the plates.
Step 2: Calculate the new capacitance with the dielectric
When a dielectric slab is inserted, the capacitance increases by a factor equal to the dielectric constant \(K\) :
\(
C^{\prime}=K \cdot C
\)
Substituting \(K=6\) :
\(
C^{\prime}=6 C
\)
Step 3: Determine the initial energy stored in the capacitor
The energy \(U\) stored in a capacitor is given by the formula:
\(
U=\frac{Q^2}{2 C}
\)
where \(Q\) is the charge on the plates.
Step 4: Calculate the new energy with the dielectric
Since the charge \(Q\) remains constant, the new energy \(U^{\prime}\) can be expressed as:
\(
U^{\prime}=\frac{Q^2}{2 C^{\prime}}
\)
Substituting \(C^{\prime}=6 C\) :
\(
U^{\prime}=\frac{Q^2}{2 \cdot 6 C}=\frac{Q^2}{12 C}
\)
Now, we can relate \(U\) to \(C\) using the original energy formula:
\(
U=\frac{Q^2}{2 C}
\)
From this, we can express \(U^{\prime}\) in terms of \(U\) :
\(
U^{\prime}=\frac{U}{6}
\)
Final Results
Thus, the new capacitance and energy after inserting the dielectric are:
New capacitance: \(C^{\prime}=6 C\)
New energy: \(U^{\prime}=\frac{U}{6}\)

Hint

(a) 

\(
\begin{aligned}
&\text { Given: }\\
&\begin{aligned}
& 8 V_{\text {tiny }}=V_{\text {big }} \\
& 8 \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& 2 \mathrm{r}=\mathrm{R} \\
& V_{\text {tiny }}=\frac{K q}{r} \\
& V_{\text {big }}=\frac{K \times 8 q}{R} \\
& V_{\text {big }}=\frac{8 K q}{2 r} \\
& V_{\text {big }}=4 V_{\text {tiny }} \\
& V_{\text {big }}=4 \times 10 \Rightarrow 40 \mathrm{~V}
\end{aligned}
\)

Hint

(a) The given circuit can be simplified as

Hint

(d) Capacitance of capacitor with oil between the plate, \(C=\frac{K \varepsilon_0 A}{d}\)
If oil is removed capacitance, \(C^{\prime}=\frac{\varepsilon_0 A}{d}=\frac{C}{K}=\frac{C}{2}\)

Hint

(b) In bringing an electron towards another electron, work has to be done (since same charges repel each other). The work done stored as electrostatic potential energy, and hence, electrostatic potential energy of the system increases.

Hint

(d) 1. Understanding the Capacitor:
A capacitor stores electrical energy in the form of electrostatic potential energy. The capacitance \(C\) is defined as the ratio of the charge \(Q\) stored on one plate to the potential difference \(V\) across the plates:
\(
V=\frac{Q}{C}
\)
2. Work Done to Charge the Capacitor:
When a small charge \(d Q\) is brought to the capacitor, the work done \(d W\) in moving this charge against the potential \(V\) is given by:
\(
d W=V \cdot d Q
\)
3. Substituting for Potential:
Since \(V=\frac{Q}{C}\), we can substitute this into the work done equation:
\(
d W=\left(\frac{Q}{C}\right) d Q
\)
4. Total Work Done (Energy Stored):
To find the total work done (or energy stored) in the capacitor as we charge it from 0 to \(Q\), we integrate \(d W\) :
\(
W=\int_0^Q d W=\int_0^Q \frac{Q}{C} d Q
\)
5. Performing the Integration:
The integral can be computed as follows:
\(
W=\frac{1}{C} \int_0^Q Q d Q=\frac{1}{C}\left(\frac{Q^2}{2}\right)_0^Q=\frac{1}{C} \cdot \frac{Q^2}{2}
\)
6. Final Expression for Energy:
Thus, the energy \(U\) stored in the capacitor becomes:
\(
U=\frac{Q^2}{2 C}
\)
7. Substituting for Charge:
Since \(Q=C V\), we can substitute this into the energy equation:
\(
U=\frac{(C V)^2}{2 C}=\frac{C V^2}{2}
\)

Hint

(c) Capacitance of capacitor (C) \(=6 \mu \mathrm{~F}=6\) \(\times 10^{-6} \mathrm{~F}\); Initial potential \(\left(V_1\right)=10 \mathrm{~V}\) and final potential \(\left(V_2\right)=20 \mathrm{~V}\).
The increase in energy \((\Delta U)\)
\(
\begin{aligned}
& =\frac{1}{2} C\left(V_2^2-V_1^2\right) \\
& =\frac{1}{2} \times\left(6 \times 10^{-6}\right) \times\left[(20)^2-(10)^2\right] \\
& =\left(3 \times 10^{-6}\right) \times 300=9 \times 10^{-4} \mathrm{~J}
\end{aligned}
\)

Hint

(d) Radii of spheres \(R_1=1 \mathrm{~cm}=1 \times 10^{-2} \mathrm{~m}\); \(R_2=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\) and charges on sphere; \(Q_1=10^{-2} \mathrm{C}\) and \(Q_2=5 \times 10^{-2} \mathrm{C}\)
\(
\text { Common potential }(V)=\frac{\text { Total charge }}{\text { Total capacity }}=\frac{Q_1+Q_2}{C_1+C_2}
\)
\(
=\frac{\left(1 \times 10^{-2}\right)+\left(5 \times 10^{-2}\right)}{4 \pi \varepsilon_0 10^{-2}+4 \pi \varepsilon_0\left(2 \times 10^{-2}\right)}=\frac{6 \times 10^{-2}}{4 \pi \varepsilon_0\left(3 \times 10^{-2}\right)}
\)
Therefore final charge on smaller sphere \(=C_1 V\)
\(
=4 \pi \varepsilon_0 \times 10^{-2} \times \frac{6 \times 10^{-2}}{4 \pi \varepsilon_0 \times 3 \times 10^{-2}}=2 \times 10^{-2} \mathrm{C}
\)

Hint

(b) To orient the dipole at any angle \(\theta\) from its initial position, work has to be done on the dipole from \(\theta=0^{\circ}\) to \(\theta\).
\(\therefore \quad\) Potential energy \(=p E(1-\cos \theta)\)

Hint

(b) Potential at the centre of the sphere \(=\) potential on the surface \(=80 \mathrm{~V}\).

Hint

\(
\begin{aligned}
&\text { (b) Charge on each plate of each capacitor }\\
&\begin{aligned}
Q & = \pm C V= \pm 25 \times 10^{-6} \times 200 \\
& = \pm 5 \times 10^{-3} \mathrm{C}
\end{aligned}
\end{aligned}
\)

Hint

(d) The energy stored in the capacitor \(=\frac{1}{2} C V^2=\frac{1}{2} \times 4 \times 10^{-6} \times(400)^2=0.32 \mathrm{~J}\) This energy will be converted into heat in the resistor.

Hint

(b) 

Using \(C_{e q}=\frac{\varepsilon_0 A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}}\)
here \(C_0=\frac{\varepsilon_0 A}{d}, t_1=\frac{3 d}{8}, t_2=\frac{d}{2}, t_3=\frac{d}{8}\)
\(
K_1=K_1, \quad K_2=\frac{K_1}{1.25} \text { and } K_3=1
\)
Given \(C_{\mathrm{eq}}=2 C_0\)
\(
\begin{aligned}
& \Rightarrow 2 C_0=\frac{\varepsilon_0 A}{\frac{3 d}{8 K_1}+\frac{d \times 1.25}{2 K_1}+\frac{d}{8}} \\
& \Rightarrow \frac{2 \varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{\frac{3 d}{8 K_1}+\frac{d}{2 K_1} \times \frac{5}{4}+\frac{d}{8}} \\
& \Rightarrow 2=\frac{1}{\frac{3}{8 K_1}+\frac{5}{8 K_1}+\frac{1}{8}} \Rightarrow K_1=\frac{8}{3}=2.66
\end{aligned}
\)

Hint

(c) Given
\(
\begin{aligned}
& |\vec{P}|=5 \times 10^{-6} \mathrm{C} \mathrm{~m} \\
& |\vec{E}|=4 \times 10^5 \mathrm{~N} / \mathrm{C} \\
& \theta_i=0^{\circ} \text { and } \theta_f=60^{\circ} \\
& \Delta U=U_f-U_i \\
& =-P E \cos \theta_f+P E \cos \theta_i \\
& =P E\left[\cos \theta_i-\cos \theta_f\right) \\
& =5 \times 10^{-6} \times 4 \times 10^5\left[1-\frac{1}{2}\right) \\
& =10 \times 10^{-6} \times 10^5=1 \mathrm{~J}
\end{aligned}
\)