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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359465 Two small charged blocks of charges $5 μ C a n d 3 μ C$ are kept on a rough surface $(μ = 0.5)$ at a separation of 0.1 $m$ . Find the separation between the two blocks when they come to rest.
supporting img

1

0.27 m

2

0.17 m

3

1.2 m

4

0.45 m

Explanation:

When the charges are released from rest they move due to electrostatic repulsion. In this process two forces, electrostatic and frictional are present. From work energy theorem we get
$W_{f} + W_{e} = 0$
$W_{f} - Δ U_{e} = 0$
$- μ m g (l - l_{0}) - k Q_{1} Q_{2} [\frac{1}{l} - \frac{1}{l_{0}}] = 0$
$k Q_{1} Q_{2} (\frac{1}{l_{0}} - \frac{1}{l}) - μ m g (l - l_{0}) = 0$
on solving we get $l = 0.27 m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359466 A hot filament liberates an electron with zero initial velocity. The anode potential is 1200 $V$ . The speed of the electron when it strikes the anode is

1

2.1 \times 10^{7} m s^{- 1}

2

2.5 \times 10^{8} m s^{- 1}

3

1.5 \times 10^{5} m s^{- 1}

4

2.5 \times 10^{6} m s^{- 1}

Explanation:

Given $V = 1200 V, (K . E)_{i} = 0$
$q V = (K . E)_{f} - (K . E)_{i}$
$\Rightarrow \frac{1}{2} m v_{f}^{2} = q V$
$\Rightarrow v_{f}^{2} = \frac{2 q V}{m} = \frac{2 \times 1.6 \times 10^{- 19} \times 1200}{9.1 \times 10^{- 31}}$
$= 4.43 \times 10^{14}$
or $v_{f} = \sqrt{4.43 \times 20^{14}} = 2.1 \times 10^{7} m s^{- 1}$

KCET - 2020

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359467 Four equal charges $q$ each are placed at four corners of a square of side $a$ each. Work done in carrying a charge $- q$ from its centre to infinity is

1 zero

2

\frac{\sqrt{2} q}{π ε_{0} a}

3

\frac{q^{2}}{2 π ε_{0} a}

4

\frac{\sqrt{2} q^{2}}{π ε_{0} a}

Explanation:

The situation is shown in figure.
supporting img

In figure, $A C = B D = \sqrt{a^{2} + a^{2}} = a \sqrt{2}$
and $O A = O B = O C = O D = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$
The potential at the centre $O$ due to given charge configuration is
$V_{O} = \frac{1}{4 π ε_{0}} [\frac{q}{O A} + \frac{q}{O B} + \frac{q}{O C} + \frac{q}{O D}]$
$= \frac{1}{4 π ε_{0}} [\begin{array}{l} \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} \\ + \frac{q}{(a / \sqrt{2})} \end{array}]$
$= \frac{1}{4 π ε_{0}} [\frac{4 q}{(a / \sqrt{2})}] = \frac{\sqrt{2} q}{π ε_{0} a}$
Work done in carrying a charge $- q$ from centre $O$ to infinity is
$W_{0 \infty} = - q (V_{\infty} - V_{0})$
$= - q (0 - \frac{\sqrt{2} q}{π ε_{0} a}) (∵ V_{\infty} = 0)$
$= \frac{\sqrt{2} q^{2}}{π ε_{0} a}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359468 In the following diagram the work done in moving an isolated point charge from point $P$ to point $A$ , $B$ and $C$ is respectively as $W_{A}, W_{B}$ and $W_{C}$ , then
supporting img

1

W_{A} < W_{B} < W_{C}

2

W_{A} = W_{B} = W_{C} = 0

3

W_{A} = W_{B} \neq W_{C}

4

W_{A} > W_{B} > W_{C}

Explanation:

According to the figure, there is no other charge. A single charge when moved in a space of no field, does not experience any force. No work is done.
$W_{A} = W_{B} = W_{C} = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359469 A charged particle $q$ is shot with speed $v$ towards another fixed charged particle $Q$ . It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$ , the closest distance of approach would be
supporting img

1

r

2

2 r

3

r / 2

4

r / 4

Explanation:

At closest distance $r$ , the whole kinetic energy of charge $q$ is converted into potential energy.
$\begin{aligned} ∴ \frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \frac{Q \cdot q}{r} \\ \Rightarrow r = \frac{1}{4 π ε_{0}} \cdot \frac{2 Q \cdot q}{m v^{2}} \end{aligned}$
In next case, $r^{'} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{m (2 v)^{2}}$
$= r^{'} = \frac{1}{4} [\frac{1}{4 π ε_{0}} \cdot \frac{2 Q q}{m v^{2}}]$
$= r^{'} = \frac{r}{4}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359465 Two small charged blocks of charges $5 μ C a n d 3 μ C$ are kept on a rough surface $(μ = 0.5)$ at a separation of 0.1 $m$ . Find the separation between the two blocks when they come to rest.
supporting img

1

0.27 m

2

0.17 m

3

1.2 m

4

0.45 m

Explanation:

When the charges are released from rest they move due to electrostatic repulsion. In this process two forces, electrostatic and frictional are present. From work energy theorem we get
$W_{f} + W_{e} = 0$
$W_{f} - Δ U_{e} = 0$
$- μ m g (l - l_{0}) - k Q_{1} Q_{2} [\frac{1}{l} - \frac{1}{l_{0}}] = 0$
$k Q_{1} Q_{2} (\frac{1}{l_{0}} - \frac{1}{l}) - μ m g (l - l_{0}) = 0$
on solving we get $l = 0.27 m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359466 A hot filament liberates an electron with zero initial velocity. The anode potential is 1200 $V$ . The speed of the electron when it strikes the anode is

1

2.1 \times 10^{7} m s^{- 1}

2

2.5 \times 10^{8} m s^{- 1}

3

1.5 \times 10^{5} m s^{- 1}

4

2.5 \times 10^{6} m s^{- 1}

Explanation:

Given $V = 1200 V, (K . E)_{i} = 0$
$q V = (K . E)_{f} - (K . E)_{i}$
$\Rightarrow \frac{1}{2} m v_{f}^{2} = q V$
$\Rightarrow v_{f}^{2} = \frac{2 q V}{m} = \frac{2 \times 1.6 \times 10^{- 19} \times 1200}{9.1 \times 10^{- 31}}$
$= 4.43 \times 10^{14}$
or $v_{f} = \sqrt{4.43 \times 20^{14}} = 2.1 \times 10^{7} m s^{- 1}$

KCET - 2020

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359467 Four equal charges $q$ each are placed at four corners of a square of side $a$ each. Work done in carrying a charge $- q$ from its centre to infinity is

1 zero

2

\frac{\sqrt{2} q}{π ε_{0} a}

3

\frac{q^{2}}{2 π ε_{0} a}

4

\frac{\sqrt{2} q^{2}}{π ε_{0} a}

Explanation:

The situation is shown in figure.
supporting img

In figure, $A C = B D = \sqrt{a^{2} + a^{2}} = a \sqrt{2}$
and $O A = O B = O C = O D = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$
The potential at the centre $O$ due to given charge configuration is
$V_{O} = \frac{1}{4 π ε_{0}} [\frac{q}{O A} + \frac{q}{O B} + \frac{q}{O C} + \frac{q}{O D}]$
$= \frac{1}{4 π ε_{0}} [\begin{array}{l} \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} \\ + \frac{q}{(a / \sqrt{2})} \end{array}]$
$= \frac{1}{4 π ε_{0}} [\frac{4 q}{(a / \sqrt{2})}] = \frac{\sqrt{2} q}{π ε_{0} a}$
Work done in carrying a charge $- q$ from centre $O$ to infinity is
$W_{0 \infty} = - q (V_{\infty} - V_{0})$
$= - q (0 - \frac{\sqrt{2} q}{π ε_{0} a}) (∵ V_{\infty} = 0)$
$= \frac{\sqrt{2} q^{2}}{π ε_{0} a}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359468 In the following diagram the work done in moving an isolated point charge from point $P$ to point $A$ , $B$ and $C$ is respectively as $W_{A}, W_{B}$ and $W_{C}$ , then
supporting img

1

W_{A} < W_{B} < W_{C}

2

W_{A} = W_{B} = W_{C} = 0

3

W_{A} = W_{B} \neq W_{C}

4

W_{A} > W_{B} > W_{C}

Explanation:

According to the figure, there is no other charge. A single charge when moved in a space of no field, does not experience any force. No work is done.
$W_{A} = W_{B} = W_{C} = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359469 A charged particle $q$ is shot with speed $v$ towards another fixed charged particle $Q$ . It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$ , the closest distance of approach would be
supporting img

1

r

2

2 r

3

r / 2

4

r / 4

Explanation:

At closest distance $r$ , the whole kinetic energy of charge $q$ is converted into potential energy.
$\begin{aligned} ∴ \frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \frac{Q \cdot q}{r} \\ \Rightarrow r = \frac{1}{4 π ε_{0}} \cdot \frac{2 Q \cdot q}{m v^{2}} \end{aligned}$
In next case, $r^{'} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{m (2 v)^{2}}$
$= r^{'} = \frac{1}{4} [\frac{1}{4 π ε_{0}} \cdot \frac{2 Q q}{m v^{2}}]$
$= r^{'} = \frac{r}{4}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359465 Two small charged blocks of charges $5 μ C a n d 3 μ C$ are kept on a rough surface $(μ = 0.5)$ at a separation of 0.1 $m$ . Find the separation between the two blocks when they come to rest.
supporting img

1

0.27 m

2

0.17 m

3

1.2 m

4

0.45 m

Explanation:

When the charges are released from rest they move due to electrostatic repulsion. In this process two forces, electrostatic and frictional are present. From work energy theorem we get
$W_{f} + W_{e} = 0$
$W_{f} - Δ U_{e} = 0$
$- μ m g (l - l_{0}) - k Q_{1} Q_{2} [\frac{1}{l} - \frac{1}{l_{0}}] = 0$
$k Q_{1} Q_{2} (\frac{1}{l_{0}} - \frac{1}{l}) - μ m g (l - l_{0}) = 0$
on solving we get $l = 0.27 m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359466 A hot filament liberates an electron with zero initial velocity. The anode potential is 1200 $V$ . The speed of the electron when it strikes the anode is

1

2.1 \times 10^{7} m s^{- 1}

2

2.5 \times 10^{8} m s^{- 1}

3

1.5 \times 10^{5} m s^{- 1}

4

2.5 \times 10^{6} m s^{- 1}

Explanation:

Given $V = 1200 V, (K . E)_{i} = 0$
$q V = (K . E)_{f} - (K . E)_{i}$
$\Rightarrow \frac{1}{2} m v_{f}^{2} = q V$
$\Rightarrow v_{f}^{2} = \frac{2 q V}{m} = \frac{2 \times 1.6 \times 10^{- 19} \times 1200}{9.1 \times 10^{- 31}}$
$= 4.43 \times 10^{14}$
or $v_{f} = \sqrt{4.43 \times 20^{14}} = 2.1 \times 10^{7} m s^{- 1}$

KCET - 2020

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359467 Four equal charges $q$ each are placed at four corners of a square of side $a$ each. Work done in carrying a charge $- q$ from its centre to infinity is

1 zero

2

\frac{\sqrt{2} q}{π ε_{0} a}

3

\frac{q^{2}}{2 π ε_{0} a}

4

\frac{\sqrt{2} q^{2}}{π ε_{0} a}

Explanation:

The situation is shown in figure.
supporting img

In figure, $A C = B D = \sqrt{a^{2} + a^{2}} = a \sqrt{2}$
and $O A = O B = O C = O D = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$
The potential at the centre $O$ due to given charge configuration is
$V_{O} = \frac{1}{4 π ε_{0}} [\frac{q}{O A} + \frac{q}{O B} + \frac{q}{O C} + \frac{q}{O D}]$
$= \frac{1}{4 π ε_{0}} [\begin{array}{l} \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} \\ + \frac{q}{(a / \sqrt{2})} \end{array}]$
$= \frac{1}{4 π ε_{0}} [\frac{4 q}{(a / \sqrt{2})}] = \frac{\sqrt{2} q}{π ε_{0} a}$
Work done in carrying a charge $- q$ from centre $O$ to infinity is
$W_{0 \infty} = - q (V_{\infty} - V_{0})$
$= - q (0 - \frac{\sqrt{2} q}{π ε_{0} a}) (∵ V_{\infty} = 0)$
$= \frac{\sqrt{2} q^{2}}{π ε_{0} a}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359468 In the following diagram the work done in moving an isolated point charge from point $P$ to point $A$ , $B$ and $C$ is respectively as $W_{A}, W_{B}$ and $W_{C}$ , then
supporting img

1

W_{A} < W_{B} < W_{C}

2

W_{A} = W_{B} = W_{C} = 0

3

W_{A} = W_{B} \neq W_{C}

4

W_{A} > W_{B} > W_{C}

Explanation:

According to the figure, there is no other charge. A single charge when moved in a space of no field, does not experience any force. No work is done.
$W_{A} = W_{B} = W_{C} = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359469 A charged particle $q$ is shot with speed $v$ towards another fixed charged particle $Q$ . It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$ , the closest distance of approach would be
supporting img

1

r

2

2 r

3

r / 2

4

r / 4

Explanation:

At closest distance $r$ , the whole kinetic energy of charge $q$ is converted into potential energy.
$\begin{aligned} ∴ \frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \frac{Q \cdot q}{r} \\ \Rightarrow r = \frac{1}{4 π ε_{0}} \cdot \frac{2 Q \cdot q}{m v^{2}} \end{aligned}$
In next case, $r^{'} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{m (2 v)^{2}}$
$= r^{'} = \frac{1}{4} [\frac{1}{4 π ε_{0}} \cdot \frac{2 Q q}{m v^{2}}]$
$= r^{'} = \frac{r}{4}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359465 Two small charged blocks of charges $5 μ C a n d 3 μ C$ are kept on a rough surface $(μ = 0.5)$ at a separation of 0.1 $m$ . Find the separation between the two blocks when they come to rest.
supporting img

1

0.27 m

2

0.17 m

3

1.2 m

4

0.45 m

Explanation:

When the charges are released from rest they move due to electrostatic repulsion. In this process two forces, electrostatic and frictional are present. From work energy theorem we get
$W_{f} + W_{e} = 0$
$W_{f} - Δ U_{e} = 0$
$- μ m g (l - l_{0}) - k Q_{1} Q_{2} [\frac{1}{l} - \frac{1}{l_{0}}] = 0$
$k Q_{1} Q_{2} (\frac{1}{l_{0}} - \frac{1}{l}) - μ m g (l - l_{0}) = 0$
on solving we get $l = 0.27 m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359466 A hot filament liberates an electron with zero initial velocity. The anode potential is 1200 $V$ . The speed of the electron when it strikes the anode is

1

2.1 \times 10^{7} m s^{- 1}

2

2.5 \times 10^{8} m s^{- 1}

3

1.5 \times 10^{5} m s^{- 1}

4

2.5 \times 10^{6} m s^{- 1}

Explanation:

Given $V = 1200 V, (K . E)_{i} = 0$
$q V = (K . E)_{f} - (K . E)_{i}$
$\Rightarrow \frac{1}{2} m v_{f}^{2} = q V$
$\Rightarrow v_{f}^{2} = \frac{2 q V}{m} = \frac{2 \times 1.6 \times 10^{- 19} \times 1200}{9.1 \times 10^{- 31}}$
$= 4.43 \times 10^{14}$
or $v_{f} = \sqrt{4.43 \times 20^{14}} = 2.1 \times 10^{7} m s^{- 1}$

KCET - 2020

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359467 Four equal charges $q$ each are placed at four corners of a square of side $a$ each. Work done in carrying a charge $- q$ from its centre to infinity is

1 zero

2

\frac{\sqrt{2} q}{π ε_{0} a}

3

\frac{q^{2}}{2 π ε_{0} a}

4

\frac{\sqrt{2} q^{2}}{π ε_{0} a}

Explanation:

The situation is shown in figure.
supporting img

In figure, $A C = B D = \sqrt{a^{2} + a^{2}} = a \sqrt{2}$
and $O A = O B = O C = O D = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$
The potential at the centre $O$ due to given charge configuration is
$V_{O} = \frac{1}{4 π ε_{0}} [\frac{q}{O A} + \frac{q}{O B} + \frac{q}{O C} + \frac{q}{O D}]$
$= \frac{1}{4 π ε_{0}} [\begin{array}{l} \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} \\ + \frac{q}{(a / \sqrt{2})} \end{array}]$
$= \frac{1}{4 π ε_{0}} [\frac{4 q}{(a / \sqrt{2})}] = \frac{\sqrt{2} q}{π ε_{0} a}$
Work done in carrying a charge $- q$ from centre $O$ to infinity is
$W_{0 \infty} = - q (V_{\infty} - V_{0})$
$= - q (0 - \frac{\sqrt{2} q}{π ε_{0} a}) (∵ V_{\infty} = 0)$
$= \frac{\sqrt{2} q^{2}}{π ε_{0} a}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359468 In the following diagram the work done in moving an isolated point charge from point $P$ to point $A$ , $B$ and $C$ is respectively as $W_{A}, W_{B}$ and $W_{C}$ , then
supporting img

1

W_{A} < W_{B} < W_{C}

2

W_{A} = W_{B} = W_{C} = 0

3

W_{A} = W_{B} \neq W_{C}

4

W_{A} > W_{B} > W_{C}

Explanation:

According to the figure, there is no other charge. A single charge when moved in a space of no field, does not experience any force. No work is done.
$W_{A} = W_{B} = W_{C} = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359469 A charged particle $q$ is shot with speed $v$ towards another fixed charged particle $Q$ . It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$ , the closest distance of approach would be
supporting img

1

r

2

2 r

3

r / 2

4

r / 4

Explanation:

At closest distance $r$ , the whole kinetic energy of charge $q$ is converted into potential energy.
$\begin{aligned} ∴ \frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \frac{Q \cdot q}{r} \\ \Rightarrow r = \frac{1}{4 π ε_{0}} \cdot \frac{2 Q \cdot q}{m v^{2}} \end{aligned}$
In next case, $r^{'} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{m (2 v)^{2}}$
$= r^{'} = \frac{1}{4} [\frac{1}{4 π ε_{0}} \cdot \frac{2 Q q}{m v^{2}}]$
$= r^{'} = \frac{r}{4}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359465 Two small charged blocks of charges $5 μ C a n d 3 μ C$ are kept on a rough surface $(μ = 0.5)$ at a separation of 0.1 $m$ . Find the separation between the two blocks when they come to rest.
supporting img

1

0.27 m

2

0.17 m

3

1.2 m

4

0.45 m

Explanation:

When the charges are released from rest they move due to electrostatic repulsion. In this process two forces, electrostatic and frictional are present. From work energy theorem we get
$W_{f} + W_{e} = 0$
$W_{f} - Δ U_{e} = 0$
$- μ m g (l - l_{0}) - k Q_{1} Q_{2} [\frac{1}{l} - \frac{1}{l_{0}}] = 0$
$k Q_{1} Q_{2} (\frac{1}{l_{0}} - \frac{1}{l}) - μ m g (l - l_{0}) = 0$
on solving we get $l = 0.27 m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359466 A hot filament liberates an electron with zero initial velocity. The anode potential is 1200 $V$ . The speed of the electron when it strikes the anode is

1

2.1 \times 10^{7} m s^{- 1}

2

2.5 \times 10^{8} m s^{- 1}

3

1.5 \times 10^{5} m s^{- 1}

4

2.5 \times 10^{6} m s^{- 1}

Explanation:

Given $V = 1200 V, (K . E)_{i} = 0$
$q V = (K . E)_{f} - (K . E)_{i}$
$\Rightarrow \frac{1}{2} m v_{f}^{2} = q V$
$\Rightarrow v_{f}^{2} = \frac{2 q V}{m} = \frac{2 \times 1.6 \times 10^{- 19} \times 1200}{9.1 \times 10^{- 31}}$
$= 4.43 \times 10^{14}$
or $v_{f} = \sqrt{4.43 \times 20^{14}} = 2.1 \times 10^{7} m s^{- 1}$

KCET - 2020

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359467 Four equal charges $q$ each are placed at four corners of a square of side $a$ each. Work done in carrying a charge $- q$ from its centre to infinity is

1 zero

2

\frac{\sqrt{2} q}{π ε_{0} a}

3

\frac{q^{2}}{2 π ε_{0} a}

4

\frac{\sqrt{2} q^{2}}{π ε_{0} a}

Explanation:

The situation is shown in figure.
supporting img

In figure, $A C = B D = \sqrt{a^{2} + a^{2}} = a \sqrt{2}$
and $O A = O B = O C = O D = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$
The potential at the centre $O$ due to given charge configuration is
$V_{O} = \frac{1}{4 π ε_{0}} [\frac{q}{O A} + \frac{q}{O B} + \frac{q}{O C} + \frac{q}{O D}]$
$= \frac{1}{4 π ε_{0}} [\begin{array}{l} \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} + \frac{q}{(a / \sqrt{2})} \\ + \frac{q}{(a / \sqrt{2})} \end{array}]$
$= \frac{1}{4 π ε_{0}} [\frac{4 q}{(a / \sqrt{2})}] = \frac{\sqrt{2} q}{π ε_{0} a}$
Work done in carrying a charge $- q$ from centre $O$ to infinity is
$W_{0 \infty} = - q (V_{\infty} - V_{0})$
$= - q (0 - \frac{\sqrt{2} q}{π ε_{0} a}) (∵ V_{\infty} = 0)$
$= \frac{\sqrt{2} q^{2}}{π ε_{0} a}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359468 In the following diagram the work done in moving an isolated point charge from point $P$ to point $A$ , $B$ and $C$ is respectively as $W_{A}, W_{B}$ and $W_{C}$ , then
supporting img

1

W_{A} < W_{B} < W_{C}

2

W_{A} = W_{B} = W_{C} = 0

3

W_{A} = W_{B} \neq W_{C}

4

W_{A} > W_{B} > W_{C}

Explanation:

According to the figure, there is no other charge. A single charge when moved in a space of no field, does not experience any force. No work is done.
$W_{A} = W_{B} = W_{C} = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359469 A charged particle $q$ is shot with speed $v$ towards another fixed charged particle $Q$ . It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$ , the closest distance of approach would be
supporting img

1

r

2

2 r

3

r / 2

4

r / 4

Explanation:

At closest distance $r$ , the whole kinetic energy of charge $q$ is converted into potential energy.
$\begin{aligned} ∴ \frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \frac{Q \cdot q}{r} \\ \Rightarrow r = \frac{1}{4 π ε_{0}} \cdot \frac{2 Q \cdot q}{m v^{2}} \end{aligned}$
In next case, $r^{'} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{m (2 v)^{2}}$
$= r^{'} = \frac{1}{4} [\frac{1}{4 π ε_{0}} \cdot \frac{2 Q q}{m v^{2}}]$
$= r^{'} = \frac{r}{4}$