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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359474 A charged $+ q$ is placed at the origin $O$ of $X$ - $Y$ axes are shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $A B$ is
supporting img

1

\frac{q Q}{4 π ε_{0}} (\frac{a - b}{a b})

2

\frac{q Q}{4 π ε_{0}} (\frac{a}{b^{2}} - \frac{1}{b})

3

\frac{q Q}{4 π ε_{0}} (\frac{b - a}{a b})

4

\frac{q Q}{4 π ε_{0}} (\frac{b}{a^{2}} - \frac{1}{b})

Explanation:

Potential at point $A$ is $V_{A} = \frac{1}{4 π ε_{0}} \frac{q}{a}$
Potential at point $B$ is $V_{B} = \frac{1}{4 π ε_{0}} \frac{q}{b}$
Work done taking a charge $Q$ from $A$ to $B$ is
$W = Q (V_{B} - V_{A}) = \frac{Q q}{4 π ε_{0}} [\frac{1}{b} - \frac{1}{a}] = \frac{q Q}{4 π ε_{0}} (\frac{a - b}{a b})$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359475 The displacement of a charge $Q$ in the electric field $\vec{E} = e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}$ is $\vec{s} = a \hat{i} + b \hat{j}$ . The work done is

1

Q (a e_{1} + b e_{2})

2

Q (\sqrt{e_{1}^{2} + e_{2}^{2})} (a + b)

3

Q \sqrt{{(a e_{1})}^{2} + {(b e_{2})}^{2}}

4

Q (e_{1} + e_{2}) \sqrt{a^{2} + b^{2}}

Explanation:

By using $W = Q (\vec{E} . \vec{S})$
$\Rightarrow W = Q [(e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}) . (a \hat{i} + b \hat{j})] = Q (e_{1} a + e_{2} b)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359476 When a positive charge is taken from lower potential to a higher potential point, then its potential energy will

1 Become zero

2 Decrease

3 Increases

4 Remain unchanged

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359477 A mass of 1 $k g$ carrying a charge of 2 $C$ is accelerated through a potential of 1 $V$ . The velocity acquired by it is

1

\sqrt{2} m s^{- 1}

2

2 m s^{- 1}

3

\frac{1}{\sqrt{2}} m s^{- 1}

4

\frac{1}{2} m s^{- 1}

Explanation:

Here, $q = 2 C, m = 1 k g, V = 1 V$
Let the velocity acquired by the mass be $v$ .
Then $\frac{1}{2} m v^{2} = q V$
or $v = \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 (2 C) (1 V)}{1 k g}} = 2 m s^{- 1}$

KCET - 2018

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359474 A charged $+ q$ is placed at the origin $O$ of $X$ - $Y$ axes are shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $A B$ is
supporting img

1

\frac{q Q}{4 π ε_{0}} (\frac{a - b}{a b})

2

\frac{q Q}{4 π ε_{0}} (\frac{a}{b^{2}} - \frac{1}{b})

3

\frac{q Q}{4 π ε_{0}} (\frac{b - a}{a b})

4

\frac{q Q}{4 π ε_{0}} (\frac{b}{a^{2}} - \frac{1}{b})

Explanation:

Potential at point $A$ is $V_{A} = \frac{1}{4 π ε_{0}} \frac{q}{a}$
Potential at point $B$ is $V_{B} = \frac{1}{4 π ε_{0}} \frac{q}{b}$
Work done taking a charge $Q$ from $A$ to $B$ is
$W = Q (V_{B} - V_{A}) = \frac{Q q}{4 π ε_{0}} [\frac{1}{b} - \frac{1}{a}] = \frac{q Q}{4 π ε_{0}} (\frac{a - b}{a b})$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359475 The displacement of a charge $Q$ in the electric field $\vec{E} = e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}$ is $\vec{s} = a \hat{i} + b \hat{j}$ . The work done is

1

Q (a e_{1} + b e_{2})

2

Q (\sqrt{e_{1}^{2} + e_{2}^{2})} (a + b)

3

Q \sqrt{{(a e_{1})}^{2} + {(b e_{2})}^{2}}

4

Q (e_{1} + e_{2}) \sqrt{a^{2} + b^{2}}

Explanation:

By using $W = Q (\vec{E} . \vec{S})$
$\Rightarrow W = Q [(e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}) . (a \hat{i} + b \hat{j})] = Q (e_{1} a + e_{2} b)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359476 When a positive charge is taken from lower potential to a higher potential point, then its potential energy will

1 Become zero

2 Decrease

3 Increases

4 Remain unchanged

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359477 A mass of 1 $k g$ carrying a charge of 2 $C$ is accelerated through a potential of 1 $V$ . The velocity acquired by it is

1

\sqrt{2} m s^{- 1}

2

2 m s^{- 1}

3

\frac{1}{\sqrt{2}} m s^{- 1}

4

\frac{1}{2} m s^{- 1}

Explanation:

Here, $q = 2 C, m = 1 k g, V = 1 V$
Let the velocity acquired by the mass be $v$ .
Then $\frac{1}{2} m v^{2} = q V$
or $v = \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 (2 C) (1 V)}{1 k g}} = 2 m s^{- 1}$

KCET - 2018

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359474 A charged $+ q$ is placed at the origin $O$ of $X$ - $Y$ axes are shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $A B$ is
supporting img

1

\frac{q Q}{4 π ε_{0}} (\frac{a - b}{a b})

2

\frac{q Q}{4 π ε_{0}} (\frac{a}{b^{2}} - \frac{1}{b})

3

\frac{q Q}{4 π ε_{0}} (\frac{b - a}{a b})

4

\frac{q Q}{4 π ε_{0}} (\frac{b}{a^{2}} - \frac{1}{b})

Explanation:

Potential at point $A$ is $V_{A} = \frac{1}{4 π ε_{0}} \frac{q}{a}$
Potential at point $B$ is $V_{B} = \frac{1}{4 π ε_{0}} \frac{q}{b}$
Work done taking a charge $Q$ from $A$ to $B$ is
$W = Q (V_{B} - V_{A}) = \frac{Q q}{4 π ε_{0}} [\frac{1}{b} - \frac{1}{a}] = \frac{q Q}{4 π ε_{0}} (\frac{a - b}{a b})$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359475 The displacement of a charge $Q$ in the electric field $\vec{E} = e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}$ is $\vec{s} = a \hat{i} + b \hat{j}$ . The work done is

1

Q (a e_{1} + b e_{2})

2

Q (\sqrt{e_{1}^{2} + e_{2}^{2})} (a + b)

3

Q \sqrt{{(a e_{1})}^{2} + {(b e_{2})}^{2}}

4

Q (e_{1} + e_{2}) \sqrt{a^{2} + b^{2}}

Explanation:

By using $W = Q (\vec{E} . \vec{S})$
$\Rightarrow W = Q [(e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}) . (a \hat{i} + b \hat{j})] = Q (e_{1} a + e_{2} b)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359476 When a positive charge is taken from lower potential to a higher potential point, then its potential energy will

1 Become zero

2 Decrease

3 Increases

4 Remain unchanged

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359477 A mass of 1 $k g$ carrying a charge of 2 $C$ is accelerated through a potential of 1 $V$ . The velocity acquired by it is

1

\sqrt{2} m s^{- 1}

2

2 m s^{- 1}

3

\frac{1}{\sqrt{2}} m s^{- 1}

4

\frac{1}{2} m s^{- 1}

Explanation:

Here, $q = 2 C, m = 1 k g, V = 1 V$
Let the velocity acquired by the mass be $v$ .
Then $\frac{1}{2} m v^{2} = q V$
or $v = \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 (2 C) (1 V)}{1 k g}} = 2 m s^{- 1}$

KCET - 2018

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359474 A charged $+ q$ is placed at the origin $O$ of $X$ - $Y$ axes are shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $A B$ is
supporting img

1

\frac{q Q}{4 π ε_{0}} (\frac{a - b}{a b})

2

\frac{q Q}{4 π ε_{0}} (\frac{a}{b^{2}} - \frac{1}{b})

3

\frac{q Q}{4 π ε_{0}} (\frac{b - a}{a b})

4

\frac{q Q}{4 π ε_{0}} (\frac{b}{a^{2}} - \frac{1}{b})

Explanation:

Potential at point $A$ is $V_{A} = \frac{1}{4 π ε_{0}} \frac{q}{a}$
Potential at point $B$ is $V_{B} = \frac{1}{4 π ε_{0}} \frac{q}{b}$
Work done taking a charge $Q$ from $A$ to $B$ is
$W = Q (V_{B} - V_{A}) = \frac{Q q}{4 π ε_{0}} [\frac{1}{b} - \frac{1}{a}] = \frac{q Q}{4 π ε_{0}} (\frac{a - b}{a b})$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359475 The displacement of a charge $Q$ in the electric field $\vec{E} = e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}$ is $\vec{s} = a \hat{i} + b \hat{j}$ . The work done is

1

Q (a e_{1} + b e_{2})

2

Q (\sqrt{e_{1}^{2} + e_{2}^{2})} (a + b)

3

Q \sqrt{{(a e_{1})}^{2} + {(b e_{2})}^{2}}

4

Q (e_{1} + e_{2}) \sqrt{a^{2} + b^{2}}

Explanation:

By using $W = Q (\vec{E} . \vec{S})$
$\Rightarrow W = Q [(e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}) . (a \hat{i} + b \hat{j})] = Q (e_{1} a + e_{2} b)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359476 When a positive charge is taken from lower potential to a higher potential point, then its potential energy will

1 Become zero

2 Decrease

3 Increases

4 Remain unchanged

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359477 A mass of 1 $k g$ carrying a charge of 2 $C$ is accelerated through a potential of 1 $V$ . The velocity acquired by it is

1

\sqrt{2} m s^{- 1}

2

2 m s^{- 1}

3

\frac{1}{\sqrt{2}} m s^{- 1}

4

\frac{1}{2} m s^{- 1}

Explanation:

Here, $q = 2 C, m = 1 k g, V = 1 V$
Let the velocity acquired by the mass be $v$ .
Then $\frac{1}{2} m v^{2} = q V$
or $v = \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 (2 C) (1 V)}{1 k g}} = 2 m s^{- 1}$

KCET - 2018

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