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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359495 The charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of an equilateral triangle of side $l$ . If the net electrostatic potential energy of the system is zero, the $Q$ is equal to

1

+ q / 2

2

z e r o

3

- q / 2

4

- q

Explanation:

supporting img
Potential energy due to charges at $A$ and $B$
$= \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l}$
Potential energy due to charges at $B$ and $C$
$= \frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Potential energy due to charges at $C$ and $A$
$\frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Total potential energy
$\frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l} + \frac{2 Q q}{4 π ε_{0} l} = 0$
$∴ q^{2} = - 2 Q q Q = - q / 2$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359496 If an electron is brought towards another electron, then the electric potential energy of the system

1 Increases

2 Decreases

3 Become zero

4 Remaining the same

Explanation:

The electron has negative charge. When an electron is approaching towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative electrostatic potential energy of system of two electrons
$U = \frac{1}{4 π ε_{0}} \frac{(- e) (- e)}{r} = \frac{1}{4 π ε_{0}} \frac{e^{2}}{r}$
Thus, as $r$ decreases, potential energy increases.

AIIMS - 2011

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359497 Two equal point charges $Q$ are fixed at $x = - a$ and $x = + a$ on $X$ -axis. Another point charge is placed at the origin. The change in electrical potential energy of $Q$ , when it is displaced by a small amount $x$ along $X$ -axis, is approximately proportional to

1

x

2

x^{2}

3

x^{3}

4

\frac{1}{x}

Explanation:

Case I Electric potential energy,
$U_{1} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{a}$
supporting img

Case II Electric potential energy,
$= U_{2} = \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x}]$

Change in potential energy
$= U_{2} - U_{1}$
$= \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x} - \frac{2}{a}]$
$= \frac{1}{4 π ε_{0}} Q q [\frac{a (a - x + a + x) - 2 (a^{2} - x^{2})}{(a^{2} - x^{2}) a}]$
For $x << a, Δ U = \frac{1}{4 π ε_{0}} Q q \frac{2 x^{2}}{a^{3}}$
$\Rightarrow Δ U = x^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359498 If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

1

E^{2}

2

E

3

E^{3}

4

1 / E^{2}

Explanation:

Electrostatic energy density
$\frac{d U}{d V} = \frac{1}{2} ε_{0} E^{2}$
$∴ \frac{d U}{d V} \propto E^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359499 A hollow conducting sphere of radius $R$ is given a charge $q$ . Another point charge $q$ is taken from centre to surface. Work done in the process is
supporting img

1

\frac{1}{4 π ε_{0}} \frac{q^{2}}{R}

2

\frac{1}{4 π ε_{0}} \frac{2 q^{2}}{R}

3

\frac{1}{4 π ε_{0}} \frac{q^{2}}{2 R}

4 zero

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359495 The charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of an equilateral triangle of side $l$ . If the net electrostatic potential energy of the system is zero, the $Q$ is equal to

1

+ q / 2

2

z e r o

3

- q / 2

4

- q

Explanation:

supporting img
Potential energy due to charges at $A$ and $B$
$= \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l}$
Potential energy due to charges at $B$ and $C$
$= \frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Potential energy due to charges at $C$ and $A$
$\frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Total potential energy
$\frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l} + \frac{2 Q q}{4 π ε_{0} l} = 0$
$∴ q^{2} = - 2 Q q Q = - q / 2$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359496 If an electron is brought towards another electron, then the electric potential energy of the system

1 Increases

2 Decreases

3 Become zero

4 Remaining the same

Explanation:

The electron has negative charge. When an electron is approaching towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative electrostatic potential energy of system of two electrons
$U = \frac{1}{4 π ε_{0}} \frac{(- e) (- e)}{r} = \frac{1}{4 π ε_{0}} \frac{e^{2}}{r}$
Thus, as $r$ decreases, potential energy increases.

AIIMS - 2011

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359497 Two equal point charges $Q$ are fixed at $x = - a$ and $x = + a$ on $X$ -axis. Another point charge is placed at the origin. The change in electrical potential energy of $Q$ , when it is displaced by a small amount $x$ along $X$ -axis, is approximately proportional to

1

x

2

x^{2}

3

x^{3}

4

\frac{1}{x}

Explanation:

Case I Electric potential energy,
$U_{1} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{a}$
supporting img

Case II Electric potential energy,
$= U_{2} = \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x}]$

Change in potential energy
$= U_{2} - U_{1}$
$= \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x} - \frac{2}{a}]$
$= \frac{1}{4 π ε_{0}} Q q [\frac{a (a - x + a + x) - 2 (a^{2} - x^{2})}{(a^{2} - x^{2}) a}]$
For $x << a, Δ U = \frac{1}{4 π ε_{0}} Q q \frac{2 x^{2}}{a^{3}}$
$\Rightarrow Δ U = x^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359498 If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

1

E^{2}

2

E

3

E^{3}

4

1 / E^{2}

Explanation:

Electrostatic energy density
$\frac{d U}{d V} = \frac{1}{2} ε_{0} E^{2}$
$∴ \frac{d U}{d V} \propto E^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359499 A hollow conducting sphere of radius $R$ is given a charge $q$ . Another point charge $q$ is taken from centre to surface. Work done in the process is
supporting img

1

\frac{1}{4 π ε_{0}} \frac{q^{2}}{R}

2

\frac{1}{4 π ε_{0}} \frac{2 q^{2}}{R}

3

\frac{1}{4 π ε_{0}} \frac{q^{2}}{2 R}

4 zero

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359495 The charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of an equilateral triangle of side $l$ . If the net electrostatic potential energy of the system is zero, the $Q$ is equal to

1

+ q / 2

2

z e r o

3

- q / 2

4

- q

Explanation:

supporting img
Potential energy due to charges at $A$ and $B$
$= \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l}$
Potential energy due to charges at $B$ and $C$
$= \frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Potential energy due to charges at $C$ and $A$
$\frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Total potential energy
$\frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l} + \frac{2 Q q}{4 π ε_{0} l} = 0$
$∴ q^{2} = - 2 Q q Q = - q / 2$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359496 If an electron is brought towards another electron, then the electric potential energy of the system

1 Increases

2 Decreases

3 Become zero

4 Remaining the same

Explanation:

The electron has negative charge. When an electron is approaching towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative electrostatic potential energy of system of two electrons
$U = \frac{1}{4 π ε_{0}} \frac{(- e) (- e)}{r} = \frac{1}{4 π ε_{0}} \frac{e^{2}}{r}$
Thus, as $r$ decreases, potential energy increases.

AIIMS - 2011

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359497 Two equal point charges $Q$ are fixed at $x = - a$ and $x = + a$ on $X$ -axis. Another point charge is placed at the origin. The change in electrical potential energy of $Q$ , when it is displaced by a small amount $x$ along $X$ -axis, is approximately proportional to

1

x

2

x^{2}

3

x^{3}

4

\frac{1}{x}

Explanation:

Case I Electric potential energy,
$U_{1} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{a}$
supporting img

Case II Electric potential energy,
$= U_{2} = \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x}]$

Change in potential energy
$= U_{2} - U_{1}$
$= \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x} - \frac{2}{a}]$
$= \frac{1}{4 π ε_{0}} Q q [\frac{a (a - x + a + x) - 2 (a^{2} - x^{2})}{(a^{2} - x^{2}) a}]$
For $x << a, Δ U = \frac{1}{4 π ε_{0}} Q q \frac{2 x^{2}}{a^{3}}$
$\Rightarrow Δ U = x^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359498 If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

1

E^{2}

2

E

3

E^{3}

4

1 / E^{2}

Explanation:

Electrostatic energy density
$\frac{d U}{d V} = \frac{1}{2} ε_{0} E^{2}$
$∴ \frac{d U}{d V} \propto E^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359499 A hollow conducting sphere of radius $R$ is given a charge $q$ . Another point charge $q$ is taken from centre to surface. Work done in the process is
supporting img

1

\frac{1}{4 π ε_{0}} \frac{q^{2}}{R}

2

\frac{1}{4 π ε_{0}} \frac{2 q^{2}}{R}

3

\frac{1}{4 π ε_{0}} \frac{q^{2}}{2 R}

4 zero

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359495 The charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of an equilateral triangle of side $l$ . If the net electrostatic potential energy of the system is zero, the $Q$ is equal to

1

+ q / 2

2

z e r o

3

- q / 2

4

- q

Explanation:

supporting img
Potential energy due to charges at $A$ and $B$
$= \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l}$
Potential energy due to charges at $B$ and $C$
$= \frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Potential energy due to charges at $C$ and $A$
$\frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Total potential energy
$\frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l} + \frac{2 Q q}{4 π ε_{0} l} = 0$
$∴ q^{2} = - 2 Q q Q = - q / 2$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359496 If an electron is brought towards another electron, then the electric potential energy of the system

1 Increases

2 Decreases

3 Become zero

4 Remaining the same

Explanation:

The electron has negative charge. When an electron is approaching towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative electrostatic potential energy of system of two electrons
$U = \frac{1}{4 π ε_{0}} \frac{(- e) (- e)}{r} = \frac{1}{4 π ε_{0}} \frac{e^{2}}{r}$
Thus, as $r$ decreases, potential energy increases.

AIIMS - 2011

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359497 Two equal point charges $Q$ are fixed at $x = - a$ and $x = + a$ on $X$ -axis. Another point charge is placed at the origin. The change in electrical potential energy of $Q$ , when it is displaced by a small amount $x$ along $X$ -axis, is approximately proportional to

1

x

2

x^{2}

3

x^{3}

4

\frac{1}{x}

Explanation:

Case I Electric potential energy,
$U_{1} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{a}$
supporting img

Case II Electric potential energy,
$= U_{2} = \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x}]$

Change in potential energy
$= U_{2} - U_{1}$
$= \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x} - \frac{2}{a}]$
$= \frac{1}{4 π ε_{0}} Q q [\frac{a (a - x + a + x) - 2 (a^{2} - x^{2})}{(a^{2} - x^{2}) a}]$
For $x << a, Δ U = \frac{1}{4 π ε_{0}} Q q \frac{2 x^{2}}{a^{3}}$
$\Rightarrow Δ U = x^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359498 If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

1

E^{2}

2

E

3

E^{3}

4

1 / E^{2}

Explanation:

Electrostatic energy density
$\frac{d U}{d V} = \frac{1}{2} ε_{0} E^{2}$
$∴ \frac{d U}{d V} \propto E^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359499 A hollow conducting sphere of radius $R$ is given a charge $q$ . Another point charge $q$ is taken from centre to surface. Work done in the process is
supporting img

1

\frac{1}{4 π ε_{0}} \frac{q^{2}}{R}

2

\frac{1}{4 π ε_{0}} \frac{2 q^{2}}{R}

3

\frac{1}{4 π ε_{0}} \frac{q^{2}}{2 R}

4 zero

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359495 The charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of an equilateral triangle of side $l$ . If the net electrostatic potential energy of the system is zero, the $Q$ is equal to

1

+ q / 2

2

z e r o

3

- q / 2

4

- q

Explanation:

supporting img
Potential energy due to charges at $A$ and $B$
$= \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l}$
Potential energy due to charges at $B$ and $C$
$= \frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Potential energy due to charges at $C$ and $A$
$\frac{1}{4 π ε_{0}} \cdot \frac{Q q}{l}$
Total potential energy
$\frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l} + \frac{2 Q q}{4 π ε_{0} l} = 0$
$∴ q^{2} = - 2 Q q Q = - q / 2$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359496 If an electron is brought towards another electron, then the electric potential energy of the system

1 Increases

2 Decreases

3 Become zero

4 Remaining the same

Explanation:

The electron has negative charge. When an electron is approaching towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative electrostatic potential energy of system of two electrons
$U = \frac{1}{4 π ε_{0}} \frac{(- e) (- e)}{r} = \frac{1}{4 π ε_{0}} \frac{e^{2}}{r}$
Thus, as $r$ decreases, potential energy increases.

AIIMS - 2011

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359497 Two equal point charges $Q$ are fixed at $x = - a$ and $x = + a$ on $X$ -axis. Another point charge is placed at the origin. The change in electrical potential energy of $Q$ , when it is displaced by a small amount $x$ along $X$ -axis, is approximately proportional to

1

x

2

x^{2}

3

x^{3}

4

\frac{1}{x}

Explanation:

Case I Electric potential energy,
$U_{1} = \frac{1}{4 π ε_{0}} \frac{2 Q q}{a}$
supporting img

Case II Electric potential energy,
$= U_{2} = \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x}]$

Change in potential energy
$= U_{2} - U_{1}$
$= \frac{1}{4 π ε_{0}} Q q [\frac{1}{a + x} + \frac{1}{a - x} - \frac{2}{a}]$
$= \frac{1}{4 π ε_{0}} Q q [\frac{a (a - x + a + x) - 2 (a^{2} - x^{2})}{(a^{2} - x^{2}) a}]$
For $x << a, Δ U = \frac{1}{4 π ε_{0}} Q q \frac{2 x^{2}}{a^{3}}$
$\Rightarrow Δ U = x^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359498 If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

1

E^{2}

2

E

3

E^{3}

4

1 / E^{2}

Explanation:

Electrostatic energy density
$\frac{d U}{d V} = \frac{1}{2} ε_{0} E^{2}$
$∴ \frac{d U}{d V} \propto E^{2}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359499 A hollow conducting sphere of radius $R$ is given a charge $q$ . Another point charge $q$ is taken from centre to surface. Work done in the process is
supporting img

1

\frac{1}{4 π ε_{0}} \frac{q^{2}}{R}

2

\frac{1}{4 π ε_{0}} \frac{2 q^{2}}{R}

3

\frac{1}{4 π ε_{0}} \frac{q^{2}}{2 R}

4 zero

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