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PHXII01:ELECTRIC CHARGES AND FIELDS

358138 Two point charges of $5 μ C and - 5 μ C$ are separated by a distance of $10 \overset{\circ}{A}$ , $A$ point $B$ is at a distance of 10 $c m$ from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at $B$ will be

1

5 N C^{- 1}

2

0.9 N C^{- 1}

3

45 N C^{- 1}

4

0.045 N C^{- 1}

Explanation:

The point lies on equatorial line of a short dipole.
$∴ E = \frac{q d}{4 π ε_{0} r^{3}} = \frac{9 \times 10^{9} \times 5 \times 10^{- 6} \times 10^{- 9}}{{(10^{- 1})}^{3}}$
$= 0.045 N C^{- 1}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358139 In the situation shown in the diagram, magnitude of $q << | Q |$ and $r >> a$ . The net force on the free charge $- q$ and net torque on it about $O$ at the instant shown are respectively $[p = 2 a Q$ is the dipole moment]
supporting img

1

- \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}, - \frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}

2

\frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}, + \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}

3

\frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}, - \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}

4

\frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}, \frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}

Explanation:

Electric field due to dipole at point $P$ is
$E = \frac{1}{4 π ε_{0}} \frac{p}{r^{3}}$ (opposite to the direction of dipole)
$\overset{―}{F} = - q \cdot \overset{―}{E}$
$\Rightarrow \overset{―}{F} = \frac{1}{4 π ε_{0}} \frac{q p}{r^{3}} \hat{i}$
Torque $\vec{τ} = \vec{r} \times \vec{F}$
$\Rightarrow \vec{τ} = - \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}$ .
Correct option is (3).

KCET - 2023

PHXII01:ELECTRIC CHARGES AND FIELDS

358140 What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line?

1

0^{\circ}

2

90^{\circ}

3

180^{\circ}

4 None of these

Explanation:

Observing $E_{n e t}$ and $p$ are in opposite directions, so angle between them is $180^{\circ}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358141 Two charges $+ 10 μ C and - 10 μ C$ are placed 20 $m m$ apart. The electric field at point $P$ , on the axis of the dipole 100 $c m$ away from its centre $O$ on the side of the positive charge is
supporting img

1

9 \times 10^{9} N C^{- 1}

2

4.2 \times 10^{6} N C^{- 1}

3

3.6 \times 10^{3} N C^{- 1}

4

3.6 \times 10^{5} N C^{- 1}

Explanation:

Here $q = \pm 20 μ C = \pm 20 \times 10^{- 6} C,$
$d = 20 m m = 20 \times 10^{- 3} m$
$r = O P = 100 c m = 1 m$
$| \vec{p} | = q \times d = 10 \times 10^{- 6} \times 20 \times 10^{- 3} = 2 \times 10^{- 7} C m$
The electric field along $B P$ ,
As $d << r,$
$\vec{E} = \frac{2 | \vec{p} |}{4 π ε_{0} r^{3}} = \frac{2 \times 2 \times 10^{- 7} \times 9 \times 10^{9}}{{(1)}^{3}}$
$= 3.6 \times 10^{3} N C^{- 1}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358138 Two point charges of $5 μ C and - 5 μ C$ are separated by a distance of $10 \overset{\circ}{A}$ , $A$ point $B$ is at a distance of 10 $c m$ from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at $B$ will be

1

5 N C^{- 1}

2

0.9 N C^{- 1}

3

45 N C^{- 1}

4

0.045 N C^{- 1}

Explanation:

The point lies on equatorial line of a short dipole.
$∴ E = \frac{q d}{4 π ε_{0} r^{3}} = \frac{9 \times 10^{9} \times 5 \times 10^{- 6} \times 10^{- 9}}{{(10^{- 1})}^{3}}$
$= 0.045 N C^{- 1}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358139 In the situation shown in the diagram, magnitude of $q << | Q |$ and $r >> a$ . The net force on the free charge $- q$ and net torque on it about $O$ at the instant shown are respectively $[p = 2 a Q$ is the dipole moment]
supporting img

1

- \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}, - \frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}

2

\frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}, + \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}

3

\frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}, - \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}

4

\frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}, \frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}

Explanation:

Electric field due to dipole at point $P$ is
$E = \frac{1}{4 π ε_{0}} \frac{p}{r^{3}}$ (opposite to the direction of dipole)
$\overset{―}{F} = - q \cdot \overset{―}{E}$
$\Rightarrow \overset{―}{F} = \frac{1}{4 π ε_{0}} \frac{q p}{r^{3}} \hat{i}$
Torque $\vec{τ} = \vec{r} \times \vec{F}$
$\Rightarrow \vec{τ} = - \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}$ .
Correct option is (3).

KCET - 2023

PHXII01:ELECTRIC CHARGES AND FIELDS

358140 What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line?

1

0^{\circ}

2

90^{\circ}

3

180^{\circ}

4 None of these

Explanation:

Observing $E_{n e t}$ and $p$ are in opposite directions, so angle between them is $180^{\circ}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358141 Two charges $+ 10 μ C and - 10 μ C$ are placed 20 $m m$ apart. The electric field at point $P$ , on the axis of the dipole 100 $c m$ away from its centre $O$ on the side of the positive charge is
supporting img

1

9 \times 10^{9} N C^{- 1}

2

4.2 \times 10^{6} N C^{- 1}

3

3.6 \times 10^{3} N C^{- 1}

4

3.6 \times 10^{5} N C^{- 1}

Explanation:

Here $q = \pm 20 μ C = \pm 20 \times 10^{- 6} C,$
$d = 20 m m = 20 \times 10^{- 3} m$
$r = O P = 100 c m = 1 m$
$| \vec{p} | = q \times d = 10 \times 10^{- 6} \times 20 \times 10^{- 3} = 2 \times 10^{- 7} C m$
The electric field along $B P$ ,
As $d << r,$
$\vec{E} = \frac{2 | \vec{p} |}{4 π ε_{0} r^{3}} = \frac{2 \times 2 \times 10^{- 7} \times 9 \times 10^{9}}{{(1)}^{3}}$
$= 3.6 \times 10^{3} N C^{- 1}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358138 Two point charges of $5 μ C and - 5 μ C$ are separated by a distance of $10 \overset{\circ}{A}$ , $A$ point $B$ is at a distance of 10 $c m$ from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at $B$ will be

1

5 N C^{- 1}

2

0.9 N C^{- 1}

3

45 N C^{- 1}

4

0.045 N C^{- 1}

Explanation:

The point lies on equatorial line of a short dipole.
$∴ E = \frac{q d}{4 π ε_{0} r^{3}} = \frac{9 \times 10^{9} \times 5 \times 10^{- 6} \times 10^{- 9}}{{(10^{- 1})}^{3}}$
$= 0.045 N C^{- 1}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358139 In the situation shown in the diagram, magnitude of $q << | Q |$ and $r >> a$ . The net force on the free charge $- q$ and net torque on it about $O$ at the instant shown are respectively $[p = 2 a Q$ is the dipole moment]
supporting img

1

- \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}, - \frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}

2

\frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}, + \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}

3

\frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}, - \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}

4

\frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}, \frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}

Explanation:

Electric field due to dipole at point $P$ is
$E = \frac{1}{4 π ε_{0}} \frac{p}{r^{3}}$ (opposite to the direction of dipole)
$\overset{―}{F} = - q \cdot \overset{―}{E}$
$\Rightarrow \overset{―}{F} = \frac{1}{4 π ε_{0}} \frac{q p}{r^{3}} \hat{i}$
Torque $\vec{τ} = \vec{r} \times \vec{F}$
$\Rightarrow \vec{τ} = - \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}$ .
Correct option is (3).

KCET - 2023

PHXII01:ELECTRIC CHARGES AND FIELDS

358140 What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line?

1

0^{\circ}

2

90^{\circ}

3

180^{\circ}

4 None of these

Explanation:

Observing $E_{n e t}$ and $p$ are in opposite directions, so angle between them is $180^{\circ}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358141 Two charges $+ 10 μ C and - 10 μ C$ are placed 20 $m m$ apart. The electric field at point $P$ , on the axis of the dipole 100 $c m$ away from its centre $O$ on the side of the positive charge is
supporting img

1

9 \times 10^{9} N C^{- 1}

2

4.2 \times 10^{6} N C^{- 1}

3

3.6 \times 10^{3} N C^{- 1}

4

3.6 \times 10^{5} N C^{- 1}

Explanation:

Here $q = \pm 20 μ C = \pm 20 \times 10^{- 6} C,$
$d = 20 m m = 20 \times 10^{- 3} m$
$r = O P = 100 c m = 1 m$
$| \vec{p} | = q \times d = 10 \times 10^{- 6} \times 20 \times 10^{- 3} = 2 \times 10^{- 7} C m$
The electric field along $B P$ ,
As $d << r,$
$\vec{E} = \frac{2 | \vec{p} |}{4 π ε_{0} r^{3}} = \frac{2 \times 2 \times 10^{- 7} \times 9 \times 10^{9}}{{(1)}^{3}}$
$= 3.6 \times 10^{3} N C^{- 1}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358138 Two point charges of $5 μ C and - 5 μ C$ are separated by a distance of $10 \overset{\circ}{A}$ , $A$ point $B$ is at a distance of 10 $c m$ from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at $B$ will be

1

5 N C^{- 1}

2

0.9 N C^{- 1}

3

45 N C^{- 1}

4

0.045 N C^{- 1}

Explanation:

The point lies on equatorial line of a short dipole.
$∴ E = \frac{q d}{4 π ε_{0} r^{3}} = \frac{9 \times 10^{9} \times 5 \times 10^{- 6} \times 10^{- 9}}{{(10^{- 1})}^{3}}$
$= 0.045 N C^{- 1}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358139 In the situation shown in the diagram, magnitude of $q << | Q |$ and $r >> a$ . The net force on the free charge $- q$ and net torque on it about $O$ at the instant shown are respectively $[p = 2 a Q$ is the dipole moment]
supporting img

1

- \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}, - \frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}

2

\frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}, + \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}

3

\frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}, - \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}

4

\frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}, \frac{1}{4 π ε_{0}} \frac{p q}{r^{3}} \hat{i}

Explanation:

Electric field due to dipole at point $P$ is
$E = \frac{1}{4 π ε_{0}} \frac{p}{r^{3}}$ (opposite to the direction of dipole)
$\overset{―}{F} = - q \cdot \overset{―}{E}$
$\Rightarrow \overset{―}{F} = \frac{1}{4 π ε_{0}} \frac{q p}{r^{3}} \hat{i}$
Torque $\vec{τ} = \vec{r} \times \vec{F}$
$\Rightarrow \vec{τ} = - \frac{1}{4 π ε_{0}} \frac{p q}{r^{2}} \hat{k}$ .
Correct option is (3).

KCET - 2023

PHXII01:ELECTRIC CHARGES AND FIELDS

358140 What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line?

1

0^{\circ}

2

90^{\circ}

3

180^{\circ}

4 None of these

Explanation:

Observing $E_{n e t}$ and $p$ are in opposite directions, so angle between them is $180^{\circ}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358141 Two charges $+ 10 μ C and - 10 μ C$ are placed 20 $m m$ apart. The electric field at point $P$ , on the axis of the dipole 100 $c m$ away from its centre $O$ on the side of the positive charge is
supporting img

1

9 \times 10^{9} N C^{- 1}

2

4.2 \times 10^{6} N C^{- 1}

3

3.6 \times 10^{3} N C^{- 1}

4

3.6 \times 10^{5} N C^{- 1}

Explanation:

Here $q = \pm 20 μ C = \pm 20 \times 10^{- 6} C,$
$d = 20 m m = 20 \times 10^{- 3} m$
$r = O P = 100 c m = 1 m$
$| \vec{p} | = q \times d = 10 \times 10^{- 6} \times 20 \times 10^{- 3} = 2 \times 10^{- 7} C m$
The electric field along $B P$ ,
As $d << r,$
$\vec{E} = \frac{2 | \vec{p} |}{4 π ε_{0} r^{3}} = \frac{2 \times 2 \times 10^{- 7} \times 9 \times 10^{9}}{{(1)}^{3}}$
$= 3.6 \times 10^{3} N C^{- 1}$
supporting img

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