Electrostatic Potential and Capacitance

To find the energy density (energy per unit volume) between the plates of the capacitor, we use the formula:

$$\text{Energy Density} (u) = \frac{1}{2} \epsilon_0 E^2$$

where:

• $\epsilon_0$​ = Permittivity of free space $= 8.85 \times 10^{-12} \, F/m$
• $E$ = Electric field inside the capacitor, given by:

$$E = \frac{V}{d}$$​

$$E = \frac{V}{d} = \frac{20V}{1 \times 10^{-6} m}$$

$$E = 2 \times 10^7 \, V/m$$

$$u = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (2 \times 10^7)^2$$

$$u = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (4 \times 10^{14})$$

$$u = \frac{1}{2} \times 3.54 \times 10^3$$

$$u = 1.77 \times 10^3 \, J/m^3$$

$\mathbf{1770}\mathbf{J}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$

So, the correct option is (A) 1770 J/m³.

$$\begin{gathered} U_T = \frac{1}{4\pi {\epsilon }_0}\left[-\frac{2q^2}{a}-\frac{2q^2}{a}+\frac{q^2}{2a}\right] \\ =\frac{1}{4\pi {\epsilon }_{0}a}\left[-4+\frac{1}{2}\right] \\ {U}_T =\frac{-7q^2}{8\pi {\epsilon }_{0}a} \end{gathered}$$

$$\begin{gathered} V=\frac{9\times {10}^9\times 4\times {10}^{-7}}{9\times {10}^{-2}} \\ V=4\times {10}^4 V \end{gathered}$$

The charge on a capacitor is given by:

$$Q = C V$$

Given:

• $C_1 = 6 \mu F$
• $V = 5V$

$$Q_1 = (6 \times 10^{-6} F) \times (5V) = 30 \mu C$$

Since $C_2$​ is initially uncharged, its initial charge is:

$$Q_2 = 0 \mu C$$

When the two capacitors are connected, charge redistributes while maintaining charge conservation:

$$Q_{\text{total}} = Q_1 + Q_2 = 30 \mu C + 0 = 30 \mu C$$

Since the capacitors are connected in parallel, they share a common voltage $V_f$:

$$V_f = \frac{Q_{\text{total}}}{C_1 + C_2}$$

$$V_f = \frac{30 \mu C}{(6 + 12) \mu F}$$V

​ $$V_f = \frac{30}{18} V = \frac{5}{3} V$$

Using $Q = CV$

1. Final charge on $C_1$:

$$Q_1' = C_1 V_f = (6 \mu F) \times \frac{5}{3} V$$

$$Q_1' = 10 \mu C$$

1. Final charge on $C_2$:

$$Q_2' = C_2 V_f = (12 \mu F) \times \frac{5}{3} V$$

$$Q_2' = 20 \mu C$$

The charges on $C_1$​ and $C_2$​ after connection are:

$$\boxed{B} \quad 10 \mu C, 20 \mu C$$

1. Initial Charge Stored:
   The charge stored in a capacitor is given by:

   $$Q = CV$$

   Substituting $C = 40 \, \mu \text{F}$ and $V = 100 \, \text{V}$:

   $$Q = 40 \times 10^{-6} \times 100 = 4 \, \text{mC}.$$

2. Capacitance with Dielectric:
   When the dielectric is introduced, the capacitance becomes:

   $$C' = K \cdot C = 2 \cdot 40 = 80 \, \mu \text{F}.$$

3. Charge with Dielectric:

   $$Q' = C' \cdot V = 80 \times 10^{-6} \cdot 100 = 8 \, \text{mC}.$$

   The extra charge stored is:

   $$\Delta Q = Q' - Q = 8 - 4 = 4 \, \text{mC}.$$

4. Change in Electrostatic Energy:
   The energy stored in a capacitor is given by:

   $$U = \frac{1}{2} CV^2.$$
   • Initial energy: $$U = \frac{1}{2} \cdot 40 \times 10^{-6} \cdot (100)^2 = 0.2 \, \text{J}.$$
   • Final energy with dielectric: $$U' = \frac{1}{2} \cdot 80 \times 10^{-6} \cdot (100)^2 = 0.4 \, \text{J}.$$

   The change in energy is:

   $$\Delta U = U' - U = 0.4 - 0.2 = 0.2 \, \text{J}.$$

Thus, the extra charge stored is $4 \, \text{mC}$ and the change in energy is $0.2\text{J}$.

The capacitance C of a parallel plate capacitor is given by the formula:

$C=\frac{{\epsilon }_{0}A}{d}$

To increase the capacitance by a factor of 10 , either the area A (which depends on $l\times b$) should be increased, or the distance d should be reduced.
Option C: $l=6cm,b=5cm,d=3\mu m$ increases the area, so the capacitance becomes 10 times the initial capacitance.
Option E: $l=5cm,b=2cm,d=1\mu m$ reduces the distance d and increases the capacitance by 10 times.
Thus, Options C and E increase the capacitance by a factor of 10 .

Analyzing the influence of multiple point charges at a distance. At large distances, the individual effects of point charges q₁​,q₂​,q_3​ on the potential can be considered to merge into a single effect similar to that of a single point charge with their cumulative charge. Therefore, the potential V at a point in space due to these charges resembles that from a single equivalent charge.
Considering the shape of equipotentials. The equipotential surfaces of a single point charge are spherical. Since at large distances the group of charges behaves like a single point charge, the equipotential surfaces thus formed will also be spherical.

• Calculate the potential difference (\(\Delta V\)):

  \[\Delta V = V_2 - V_1 = (+5 \text{ V}) - (-5 \text{ V}) = 10 \text{ V}\]

• Calculate the work done (\(W_{\text{ext}}\)):

  \[W_{\text{ext}} = q \Delta V\] \[W_{\text{ext}} = (1.6 \times 10^{-19} \text{ C}) \times (10 \text{ V})\] \[W_{\text{ext}} = 16 \times 10^{-19} \text{ J} = 1.6 \times 10^{-18} \text{ J}\]

• Calculate the potential difference (\(\Delta V\)):

  \[\Delta V = V_2 - V_1 = (+5 \text{ V}) - (-5 \text{ V}) = 10 \text{ V}\]

• Calculate the work done (\(W_{\text{ext}}\)):

  \[W_{\text{ext}} = q \Delta V\] \[W_{\text{ext}} = (1.6 \times 10^{-19} \text{ C}) \times (10 \text{ V})\] \[W_{\text{ext}} = 16 \times 10^{-19} \text{ J} = 1.6 \times 10^{-18} \text{ J}\]

The electrostatic potential due to an electric dipole at a distance ' \(r\) ' varies as \(\frac{1}{r^2}\)

$$\begin{gathered} U=\frac{k{q}_1{q}_2}{\gamma } \\ U=\frac{-9\times {10}^9\times 4\times {10}^{-6}\times 7\times {10}^{-6}\times 100}{14} \\ U=-18\times {10}^{-1}J \\ U=-1.8J \end{gathered}$$

Energy per unit volume is given by:

$\frac{U}{V}=\frac{1}{2}{\epsilon }_0{E}^2$

Since volume V of the capacitor is $A\times d$, the total energy stored in the capacitor is:

$U=\frac{1}{2}{\epsilon }_0{E}^2\times Ad$

Battery is disconnected, so \(\mathbf{Q = \text{constant}}\).

Capacitance (C): \(C = \frac{\epsilon_0 A}{d}\)
As \(d\) (separation) increases, \(C\) decreases.
Potential Difference (V): \(V = \frac{Q}{C}\)
As \(Q\) is constant and \(C\) decreases, \(V\) increases.
Electrostatic Energy (U): \(U = \frac{Q^2}{2C}\)
As \(Q\) is constant and \(C\) decreases, \(U\) increases.
Final Answer:
D. the electrostatic energy stored in the capacitor increases.

For 10 Capacitors in parallel $C_{eq} =10C$

\begin{equation}
\begin{aligned}
&=10 \times 10^{-6}=1 \times 10^{-5} \mathrm{~F}\\
&\text { Energy stored: } U=\frac{1}{2} C_{\text {eq }} V^2\\
&\begin{aligned}
& U=\frac{1}{2} \times 1 \times 10^{-5} \times(100)^2 \\
& \begin{aligned}
U & =\frac{1}{2} \times 1 \times 10^{-5} \times 10^4=0.5 \times 10^{-1} \\
& =5 \times 10^{-2} \mathrm{~J}
\end{aligned}
\end{aligned}
\end{aligned}
\end{equation}

Analyzing the influence of multiple point charges at a distance. At large distances, the individual effects of point charges q₁​,q₂​,q_3​ on the potential can be considered to merge into a single effect similar to that of a single point charge with their cumulative charge. Therefore, the potential V at a point in space due to these charges resembles that from a single equivalent charge.
Considering the shape of equipotentials. The equipotential surfaces of a single point charge are spherical. Since at large distances the group of charges behaves like a single point charge, the equipotential surfaces thus formed will also be spherical.

The capacitance of a capacitor depends on the geometry of the capaciitor, which includes the size and shape of the plates and the distance between them. The capacitance is independent of the charge (q) or potential difference (V)

$$\begin{gathered} U_T = \frac{1}{4\pi {\epsilon }_0}\left[-\frac{2q^2}{a}-\frac{2q^2}{a}+\frac{q^2}{2a}\right] \\ =\frac{1}{4\pi {\epsilon }_{0}a}\left[-4+\frac{1}{2}\right] \\ {U}_T =\frac{-7q^2}{8\pi {\epsilon }_{0}a} \end{gathered}$$

The capacitance C of a parallel plate capacitor is given by the formula:

$C=\frac{{\epsilon }_{0}A}{d}$

To increase the capacitance by a factor of 10 , either the area A (which depends on $l\times b$) should be increased, or the distance d should be reduced.
Option C: $l=6cm,b=5cm,d=3\mu m$ increases the area, so the capacitance becomes 10 times the initial capacitance.
Option E: $l=5cm,b=2cm,d=1\mu m$ reduces the distance d and increases the capacitance by 10 times.
Thus, Options C and E increase the capacitance by a factor of 10 .

$$\begin{gathered} {\mathrm{E}}_{\mathrm{axial}} = \frac{2\mathrm{kP}}{{\mathrm{r}}^3}={\mathrm{E}}_{\mathrm{o}} \\ {\mathrm{E}}_{\mathrm{equitorial} }=\frac{\mathrm{kP}}{{\left(2\mathrm{r}\right)}^3}=\frac{\mathrm{kP}}{8{\mathrm{r}}^3}=\frac{{\mathrm{E}}_{\mathrm{o}}}{16} \\ \mathrm{V}=\frac{\mathrm{kP} \mathrm{cos\theta }}{{\mathrm{r}}^2} \\ {\mathrm{V}}_{\mathrm{axial}}=\frac{\mathrm{kP} \cos 0}{{\mathrm{r}}^2}=\frac{\mathrm{kP}}{{\mathrm{r}}^2} \\ {\mathrm{V}}_{\mathrm{equitorial}}=\frac{\mathrm{kP} \cos 90}{{\left(2\mathrm{r}\right)}^2}=0 \end{gathered}$$

Analyzing the influence of multiple point charges at a distance. At large distances, the individual effects of point charges q₁​,q₂​,q_3​ on the potential can be considered to merge into a single effect similar to that of a single point charge with their cumulative charge. Therefore, the potential V at a point in space due to these charges resembles that from a single equivalent charge.
Considering the shape of equipotentials. The equipotential surfaces of a single point charge are spherical. Since at large distances the group of charges behaves like a single point charge, the equipotential surfaces thus formed will also be spherical.

Electric potential due to line charge at a distance r is

$$\begin{gathered} V\left(x\right)=V_0-2K\lambda \ln \left(\frac{x}{x_0}\right) \\ V\left(x\right)=\left(V_0+2K\lambda \ln x_0\right)-2K\lambda \ln x \\ =C-2K\lambda \ln x \\ V(x,y,z)=V_1+V_2+V_3 \\ =C_1+C_2+C_3-2K\lambda \ln {\left(x_2+y^2\right)}^{1/2} \end{gathered}$$

$2K\lambda \ln {\left(y_2+z_2\right)}^{1/2}-2K\lambda \ln {\left(x^2+z^2\right)}^{1/2}$

$=C^{\text{'}}-2K\lambda \ln \left[\left(x^2+y^2\right)·\left(y^2+z^2\right)·\left(z^2+x^2\right)\right]$

$\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=e^{C-V}$

For equipotential surface v = Constant

$\Rightarrow \left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=\text{ Constant. }$

Since, Volume remain constant

\begin{equation}
\begin{aligned}
& 64 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \\
& R^3=r^3 \times 64 \\
& R=4 r \\
& \sigma_{\text {bigger }}=\frac{5 \times 64}{4 \pi R^2}
\end{aligned}
\end{equation}

\begin{equation}
\begin{aligned}
& \sigma_{\text {small drop }}=\frac{5}{4 \pi r^2} \\
& \frac{\sigma_{\text {bigger }}}{\sigma_{\text {small drop }}}=\frac{5 \times 64 \times 4 \pi r^2}{4 \pi R^2 \times 5} =64\left(\frac{r}{R}\right)^2=64\left(\frac{r}{4 r}\right)^2=\frac{64}{16}=4: 1
\end{aligned}
\end{equation}

$$\begin{gathered} =W=PE\left(\cos {\theta }_1-\cos {\theta }_2\right) \\ P=4\times {10}^{-6}\times \frac{18}{100} \\ P=72\times {10}^{-8}cm \\ W=72\times {10}^{-8}\times 2\times {10}^4 \\ W=14.4mJ \end{gathered}$$

Battery is disconnected, so \(\mathbf{Q = \text{constant}}\).

Capacitance (C): \(C = \frac{\epsilon_0 A}{d}\)
As \(d\) (separation) increases, \(C\) decreases.
Potential Difference (V): \(V = \frac{Q}{C}\)
As \(Q\) is constant and \(C\) decreases, \(V\) increases.
Electrostatic Energy (U): \(U = \frac{Q^2}{2C}\)
As \(Q\) is constant and \(C\) decreases, \(U\) increases.
Final Answer:
D. the electrostatic energy stored in the capacitor increases.

For 10 Capacitors in parallel $C_{eq} =10C$

\begin{equation}
\begin{aligned}
&=10 \times 10^{-6}=1 \times 10^{-5} \mathrm{~F}\\
&\text { Energy stored: } U=\frac{1}{2} C_{\text {eq }} V^2\\
&\begin{aligned}
& U=\frac{1}{2} \times 1 \times 10^{-5} \times(100)^2 \\
& \begin{aligned}
U & =\frac{1}{2} \times 1 \times 10^{-5} \times 10^4=0.5 \times 10^{-1} \\
& =5 \times 10^{-2} \mathrm{~J}
\end{aligned}
\end{aligned}
\end{aligned}
\end{equation}

For 10 Capacitors in parallel $C_{eq} =10C$

\begin{equation}
\begin{aligned}
&=10 \times 10^{-6}=1 \times 10^{-5} \mathrm{~F}\\
&\text { Energy stored: } U=\frac{1}{2} C_{\text {eq }} V^2\\
&\begin{aligned}
& U=\frac{1}{2} \times 1 \times 10^{-5} \times(100)^2 \\
& \begin{aligned}
U & =\frac{1}{2} \times 1 \times 10^{-5} \times 10^4=0.5 \times 10^{-1} \\
& =5 \times 10^{-2} \mathrm{~J}
\end{aligned}
\end{aligned}
\end{aligned}
\end{equation}

$$\begin{gathered} U_T = \frac{1}{4\pi {\epsilon }_0}\left[-\frac{2q^2}{a}-\frac{2q^2}{a}+\frac{q^2}{2a}\right] \\ =\frac{1}{4\pi {\epsilon }_{0}a}\left[-4+\frac{1}{2}\right] \\ {U}_T =\frac{-7q^2}{8\pi {\epsilon }_{0}a} \end{gathered}$$

By the volume conservation

$$\begin{gathered} 8 \times \frac{4}{3}{\mathrm{\pi r}}^3 = \frac{4}{3}{\mathrm{\pi R}}^3 \\ \mathrm{R} = 2 \mathrm{r} \\ \mathrm{By} \mathrm{conservation} \mathrm{of} \mathrm{charge} \\ \mathrm{Q} = 8 \mathrm{q} \\ \mathrm{Now}, \\ \mathrm{V} = \frac{\mathrm{k} \mathrm{q}}{\mathrm{r}}.....\left(1\right) \\ \mathrm{V}\text{'} = \frac{\mathrm{k} \mathrm{Q}}{\mathrm{R}}=\frac{\mathrm{K} 8 \mathrm{q} }{2 \mathrm{r}}=4\mathrm{V} \\ 20 = 4 \mathrm{V} \\ \mathrm{V} = 5 \mathrm{V} \end{gathered}$$

The kinetic energy (\(KE\)) gained by a charged particle when accelerated through a potential difference (\(V\)) is given by the formula:
\(KE = qV\)
where \(q\) is the charge of the particle.

For particle A with charge \(q\) and potential difference \(V_A\):
\(KE_A = qV_A\)
For particle B with charge \(4q\) and potential difference \(V_B\):
\(KE_B = 4qV_B\)
Since both particles attain the same kinetic energy, we can equate the expressions:
\(qV_A = 4qV_B\)

We can divide both sides of the equation by \(q\) (assuming \(q \neq 0\)):
\(V_A = 4V_B\)

Now, we can express the ratio of the potential differences:
\(\frac{V_A}{V_B} = 4\)

The value of \(\left(\frac{V_A}{V_B}\right)\) is 4.

Thus, the answer is (D) 4.

Here, the line charge is placed aling the Y axis, so the equipotential surfaces are cocentric cylinders, and their crossections are circles in X Z plane. So the equation of locus,

${\mathrm{x}}^2 + {\mathrm{z}}^2 = \mathrm{Constant}$

The kinetic energy (\(KE\)) gained by a charged particle when accelerated through a potential difference (\(V\)) is given by the formula:
\(KE = qV\)
where \(q\) is the charge of the particle.

For particle A with charge \(q\) and potential difference \(V_A\):
\(KE_A = qV_A\)
For particle B with charge \(4q\) and potential difference \(V_B\):
\(KE_B = 4qV_B\)
Since both particles attain the same kinetic energy, we can equate the expressions:
\(qV_A = 4qV_B\)

We can divide both sides of the equation by \(q\) (assuming \(q \neq 0\)):
\(V_A = 4V_B\)

Now, we can express the ratio of the potential differences:
\(\frac{V_A}{V_B} = 4\)

The value of \(\left(\frac{V_A}{V_B}\right)\) is 4.

Thus, the answer is (D) 4.

The formula for electrostatic potential energy between two point charges is:

$$U = K \frac{q_1 q_2}{r}$$​​

Where:

• $K=9\times 10^9{\text{N}\text{}\text{m}}^2\mathrm{/}{\text{C}}^2$,
• $q_1 = +7 \, \text{C}$, $q_2 = -4 \, \text{C}$
• $r = \text{distance between the charges}$

The charges are located at $(-7, 0, 0)$ and $(7, 0, 0)$, so the distance $r$ is:

$$r = |7 - (-7)| = 14 \, \text{m}$$

Substitute the values into the formula:

$$U = \left( 9 \times 10^9 \right) \cdot \frac{(+7)(-4)}{14}$$​

Simplify:

$$U = \left( 9 \times 10^9 \right) \cdot \frac{-28}{14}$$

$$U = \left( 9 \times 10^9 \right) \cdot (-2) = -18 \times 10^9 \, \text{J}$$

Correct answer: (b) $-18 \times 10^9 \, \text{J}$.

1. Capacitance Formula: The capacitance of a parallel-plate capacitor is given by:

   $$C = \epsilon_0 \frac{A}{d},$$

   where:

   • $A = l \times b$ is the plate area,
   • $d$ is the separation between the plates,
   • $\epsilon_0$ is the permittivity of free space.

2. Initial Capacitance: For the given dimensions:

   • $l = 3 \, \text{cm}, b = 1 \, \text{cm}, d = 3 \, \mu\text{m}$
   $$C_{\text{initial}} = \epsilon_0 \frac{(3 \, \text{cm}) \times (1 \, \text{cm})}{3 \, \mu\text{m}} = \epsilon_0 \frac{3}{3} = \epsilon_0.$$

3. Condition for Increasing Capacitance by a Factor of 10: To increase capacitance by a factor of 10:

   $$C_{\text{new}} = 10 \times C_{\text{initial}} = 10 \epsilon_0.$$

   This requires:

   $$\frac{l \cdot b}{d} = 10 \times \frac{3 \, \text{cm} \cdot 1 \, \text{cm}}{3 \, \mu\text{m}}.$$

4. Analyze Each Option:

   • (A): $l = 6 \, \text{cm}, b = 5 \, \text{cm}, d = 3 \, \mu\text{m}$

     $$\frac{l \cdot b}{d} = \frac{6 \cdot 5}{3} = 10 \, \text{times the initial value}.$$

     Satisfies the condition.

   • (B): $l = 5 \, \text{cm}, b = 2 \, \text{cm}, d = 1 \, \mu\text{m}$

     $$\frac{l \cdot b}{d} = \frac{5 \cdot 2}{1} = 10 \, \text{times the initial value}.$$

     Satisfies the condition.

   • (C): $l = 5 \, \text{cm}, b = 1 \, \text{cm}, d = 30 \, \mu\text{m}$

     $$\frac{l \cdot b}{d} = \frac{5 \cdot 1}{30} = \frac{1}{6} \, \text{times the initial value}.$$

     Does not satisfy the condition.

   • (D): $l = 1 \, \text{cm}, b = 1 \, \text{cm}, d = 30 \, \mu\text{m}$

     $$\frac{l \cdot b}{d} = \frac{1 \cdot 1}{30} = \frac{1}{30} \, \text{times the initial value}.$$

     Does not satisfy the condition.

5. Correct Answer: The capacitance increases by a factor of 10 for configurations A and B.

Answer: (1) A, B

Here, the line charge is placed aling the Y axis, so the equipotential surfaces are cocentric cylinders, and their crossections are circles in X Z plane. So the equation of locus,

${\mathrm{x}}^2 + {\mathrm{z}}^2 = \mathrm{Constant}$

The charge on a capacitor is given by:

$$Q = C V$$

Given:

• $C_1 = 6 \mu F$
• $V = 5V$

$$Q_1 = (6 \times 10^{-6} F) \times (5V) = 30 \mu C$$

Since $C_2$​ is initially uncharged, its initial charge is:

$$Q_2 = 0 \mu C$$

When the two capacitors are connected, charge redistributes while maintaining charge conservation:

$$Q_{\text{total}} = Q_1 + Q_2 = 30 \mu C + 0 = 30 \mu C$$

Since the capacitors are connected in parallel, they share a common voltage $V_f$:

$$V_f = \frac{Q_{\text{total}}}{C_1 + C_2}$$

$$V_f = \frac{30 \mu C}{(6 + 12) \mu F}$$V

​ $$V_f = \frac{30}{18} V = \frac{5}{3} V$$

Using $Q = CV$

1. Final charge on $C_1$:

$$Q_1' = C_1 V_f = (6 \mu F) \times \frac{5}{3} V$$

$$Q_1' = 10 \mu C$$

1. Final charge on $C_2$:

$$Q_2' = C_2 V_f = (12 \mu F) \times \frac{5}{3} V$$

$$Q_2' = 20 \mu C$$

The charges on $C_1$​ and $C_2$​ after connection are:

$$\boxed{B} \quad 10 \mu C, 20 \mu C$$

The kinetic energy (\(KE\)) gained by a charged particle when accelerated through a potential difference (\(V\)) is given by the formula:
\(KE = qV\)
where \(q\) is the charge of the particle.

For particle A with charge \(q\) and potential difference \(V_A\):
\(KE_A = qV_A\)
For particle B with charge \(4q\) and potential difference \(V_B\):
\(KE_B = 4qV_B\)
Since both particles attain the same kinetic energy, we can equate the expressions:
\(qV_A = 4qV_B\)

We can divide both sides of the equation by \(q\) (assuming \(q \neq 0\)):
\(V_A = 4V_B\)

Now, we can express the ratio of the potential differences:
\(\frac{V_A}{V_B} = 4\)

The value of \(\left(\frac{V_A}{V_B}\right)\) is 4.

Thus, the answer is (D) 4.

• Calculate the potential difference (\(\Delta V\)):

  \[\Delta V = V_2 - V_1 = (+5 \text{ V}) - (-5 \text{ V}) = 10 \text{ V}\]

• Calculate the work done (\(W_{\text{ext}}\)):

  \[W_{\text{ext}} = q \Delta V\] \[W_{\text{ext}} = (1.6 \times 10^{-19} \text{ C}) \times (10 \text{ V})\] \[W_{\text{ext}} = 16 \times 10^{-19} \text{ J} = 1.6 \times 10^{-18} \text{ J}\]

• The capacitance of the parallel plate capacitor is: $$C = \epsilon_0 \frac{A}{d}.$$
• Energy stored in a capacitor: $$U = \frac{1}{2} C V^2.$$
• Substituting $V = E d$ and $C = \epsilon_0 \frac{A}{d}$​:
• $$U = \frac{1}{2} \epsilon_0 \frac{A}{d} (E d)^2 = \frac{1}{2} E^2 A \epsilon_0 d.$$

Thus, the energy stored is $\frac{1}{2} E^2 A \epsilon_0 d$

Battery is disconnected, so \(\mathbf{Q = \text{constant}}\).

Capacitance (C): \(C = \frac{\epsilon_0 A}{d}\)
As \(d\) (separation) increases, \(C\) decreases.
Potential Difference (V): \(V = \frac{Q}{C}\)
As \(Q\) is constant and \(C\) decreases, \(V\) increases.
Electrostatic Energy (U): \(U = \frac{Q^2}{2C}\)
As \(Q\) is constant and \(C\) decreases, \(U\) increases.
Final Answer:
D. the electrostatic energy stored in the capacitor increases.

The kinetic energy (\(KE\)) gained by a charged particle when accelerated through a potential difference (\(V\)) is given by the formula:
\(KE = qV\)
where \(q\) is the charge of the particle.

For particle A with charge \(q\) and potential difference \(V_A\):
\(KE_A = qV_A\)
For particle B with charge \(4q\) and potential difference \(V_B\):
\(KE_B = 4qV_B\)
Since both particles attain the same kinetic energy, we can equate the expressions:
\(qV_A = 4qV_B\)

We can divide both sides of the equation by \(q\) (assuming \(q \neq 0\)):
\(V_A = 4V_B\)

Now, we can express the ratio of the potential differences:
\(\frac{V_A}{V_B} = 4\)

The value of \(\left(\frac{V_A}{V_B}\right)\) is 4.

Thus, the answer is (D) 4.