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PHXII01:ELECTRIC CHARGES AND FIELDS

358070 Two spheres carrying charges $+ 6 μ C$ and $+ 9 μ C$ , separated by a distance $d$ , experience a force of repulsion $F$ . When a charge of $- 3 μ C$ is given to both the spheres and kept at the same distance as before, the new force of repulsion is

1

3 F

2

F / 9

3

F

4

F / 3

Explanation:

supporting img
Let us call spheres as $A$ and $B$ .
Accroding to Coulomb's law, the force of repulsion between $A$ and $B$ separated by a distance $d$ is
$F = \frac{1}{4 π ε_{0}} \frac{(+ 6 μ C) (+ 9 μ C)}{d^{2}} (1)$
When a charge of $- 3 μ C$ is given to both the spheres, then charge on $A = + 6 μ C - 3 μ C = + 3 μ C$ and on $B = + 9 μ C - 3 μ C = 6 μ C$
Again by Columbo's law, then new force of repulsion between $A$ and $B$ separated by the same distance is
$F^{'} = \frac{1}{4 π ε_{0}} \frac{(+ 3 μ C) (+ 6 μ C)}{d^{2}} (2)$
Dividing equ (2) eqn (1), we get
$\frac{F^{'}}{F} = \frac{\frac{1}{4 π ε_{0}} \frac{(+ 3 μ C) (+ 6 μ C)}{d^{2}}}{\frac{1}{4 π ε_{0}} \frac{(+ 6 μ C) (+ 9 μ C)}{d^{2}}}$
$= \frac{(+ 3 μ C) (+ 6 μ C)}{(+ 6 μ C) (+ 9 μ C)} = \frac{1}{3}$ or $F^{'} = \frac{F}{3}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358071 Assertion :
Two charges $q_{1}$ and $q_{2}$ , are placed at separation $r$ . Then, magnitude of force $(F)$ on each charge is given by coulomb law.
Reason :
Now a third charge $q_{3}$ is placed near $q_{1}$ and $q_{2}$ Then force on $q_{1}$ , due to $q_{2}$ , remains $F$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII01:ELECTRIC CHARGES AND FIELDS

358072 When a brass plate is introduced between two charges, the force between the charges

1 Decreases

2 Increases

3 Remains same

4 Becomes zero

Explanation:

Force becomes zero after inserting the slab as $k = \infty$

PHXII01:ELECTRIC CHARGES AND FIELDS

358073 Two small spheres each having the charge $+ Q$ are suspended by insulating threads of length $L$ from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between two suspensions and the tension in each will be

1

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{{(2 L)}^{2}}

2

90^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{L^{2}}

3

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{2 L^{2}}

4

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{L^{2}}

Explanation:

In the absence of gravitational force, the only force acts on the spheres is electrostatic repulsion and so the angle between two suspension becomes $180^{0}$ . So force between the spheres
$F = \frac{1}{4 π \in_{0}} \frac{Q^{2}}{{(2 L)}^{2}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358070 Two spheres carrying charges $+ 6 μ C$ and $+ 9 μ C$ , separated by a distance $d$ , experience a force of repulsion $F$ . When a charge of $- 3 μ C$ is given to both the spheres and kept at the same distance as before, the new force of repulsion is

1

3 F

2

F / 9

3

F

4

F / 3

Explanation:

supporting img
Let us call spheres as $A$ and $B$ .
Accroding to Coulomb's law, the force of repulsion between $A$ and $B$ separated by a distance $d$ is
$F = \frac{1}{4 π ε_{0}} \frac{(+ 6 μ C) (+ 9 μ C)}{d^{2}} (1)$
When a charge of $- 3 μ C$ is given to both the spheres, then charge on $A = + 6 μ C - 3 μ C = + 3 μ C$ and on $B = + 9 μ C - 3 μ C = 6 μ C$
Again by Columbo's law, then new force of repulsion between $A$ and $B$ separated by the same distance is
$F^{'} = \frac{1}{4 π ε_{0}} \frac{(+ 3 μ C) (+ 6 μ C)}{d^{2}} (2)$
Dividing equ (2) eqn (1), we get
$\frac{F^{'}}{F} = \frac{\frac{1}{4 π ε_{0}} \frac{(+ 3 μ C) (+ 6 μ C)}{d^{2}}}{\frac{1}{4 π ε_{0}} \frac{(+ 6 μ C) (+ 9 μ C)}{d^{2}}}$
$= \frac{(+ 3 μ C) (+ 6 μ C)}{(+ 6 μ C) (+ 9 μ C)} = \frac{1}{3}$ or $F^{'} = \frac{F}{3}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358071 Assertion :
Two charges $q_{1}$ and $q_{2}$ , are placed at separation $r$ . Then, magnitude of force $(F)$ on each charge is given by coulomb law.
Reason :
Now a third charge $q_{3}$ is placed near $q_{1}$ and $q_{2}$ Then force on $q_{1}$ , due to $q_{2}$ , remains $F$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII01:ELECTRIC CHARGES AND FIELDS

358072 When a brass plate is introduced between two charges, the force between the charges

1 Decreases

2 Increases

3 Remains same

4 Becomes zero

Explanation:

Force becomes zero after inserting the slab as $k = \infty$

PHXII01:ELECTRIC CHARGES AND FIELDS

358073 Two small spheres each having the charge $+ Q$ are suspended by insulating threads of length $L$ from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between two suspensions and the tension in each will be

1

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{{(2 L)}^{2}}

2

90^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{L^{2}}

3

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{2 L^{2}}

4

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{L^{2}}

Explanation:

In the absence of gravitational force, the only force acts on the spheres is electrostatic repulsion and so the angle between two suspension becomes $180^{0}$ . So force between the spheres
$F = \frac{1}{4 π \in_{0}} \frac{Q^{2}}{{(2 L)}^{2}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358070 Two spheres carrying charges $+ 6 μ C$ and $+ 9 μ C$ , separated by a distance $d$ , experience a force of repulsion $F$ . When a charge of $- 3 μ C$ is given to both the spheres and kept at the same distance as before, the new force of repulsion is

1

3 F

2

F / 9

3

F

4

F / 3

Explanation:

supporting img
Let us call spheres as $A$ and $B$ .
Accroding to Coulomb's law, the force of repulsion between $A$ and $B$ separated by a distance $d$ is
$F = \frac{1}{4 π ε_{0}} \frac{(+ 6 μ C) (+ 9 μ C)}{d^{2}} (1)$
When a charge of $- 3 μ C$ is given to both the spheres, then charge on $A = + 6 μ C - 3 μ C = + 3 μ C$ and on $B = + 9 μ C - 3 μ C = 6 μ C$
Again by Columbo's law, then new force of repulsion between $A$ and $B$ separated by the same distance is
$F^{'} = \frac{1}{4 π ε_{0}} \frac{(+ 3 μ C) (+ 6 μ C)}{d^{2}} (2)$
Dividing equ (2) eqn (1), we get
$\frac{F^{'}}{F} = \frac{\frac{1}{4 π ε_{0}} \frac{(+ 3 μ C) (+ 6 μ C)}{d^{2}}}{\frac{1}{4 π ε_{0}} \frac{(+ 6 μ C) (+ 9 μ C)}{d^{2}}}$
$= \frac{(+ 3 μ C) (+ 6 μ C)}{(+ 6 μ C) (+ 9 μ C)} = \frac{1}{3}$ or $F^{'} = \frac{F}{3}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358071 Assertion :
Two charges $q_{1}$ and $q_{2}$ , are placed at separation $r$ . Then, magnitude of force $(F)$ on each charge is given by coulomb law.
Reason :
Now a third charge $q_{3}$ is placed near $q_{1}$ and $q_{2}$ Then force on $q_{1}$ , due to $q_{2}$ , remains $F$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII01:ELECTRIC CHARGES AND FIELDS

358072 When a brass plate is introduced between two charges, the force between the charges

1 Decreases

2 Increases

3 Remains same

4 Becomes zero

Explanation:

Force becomes zero after inserting the slab as $k = \infty$

PHXII01:ELECTRIC CHARGES AND FIELDS

358073 Two small spheres each having the charge $+ Q$ are suspended by insulating threads of length $L$ from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between two suspensions and the tension in each will be

1

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{{(2 L)}^{2}}

2

90^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{L^{2}}

3

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{2 L^{2}}

4

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{L^{2}}

Explanation:

In the absence of gravitational force, the only force acts on the spheres is electrostatic repulsion and so the angle between two suspension becomes $180^{0}$ . So force between the spheres
$F = \frac{1}{4 π \in_{0}} \frac{Q^{2}}{{(2 L)}^{2}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358070 Two spheres carrying charges $+ 6 μ C$ and $+ 9 μ C$ , separated by a distance $d$ , experience a force of repulsion $F$ . When a charge of $- 3 μ C$ is given to both the spheres and kept at the same distance as before, the new force of repulsion is

1

3 F

2

F / 9

3

F

4

F / 3

Explanation:

supporting img
Let us call spheres as $A$ and $B$ .
Accroding to Coulomb's law, the force of repulsion between $A$ and $B$ separated by a distance $d$ is
$F = \frac{1}{4 π ε_{0}} \frac{(+ 6 μ C) (+ 9 μ C)}{d^{2}} (1)$
When a charge of $- 3 μ C$ is given to both the spheres, then charge on $A = + 6 μ C - 3 μ C = + 3 μ C$ and on $B = + 9 μ C - 3 μ C = 6 μ C$
Again by Columbo's law, then new force of repulsion between $A$ and $B$ separated by the same distance is
$F^{'} = \frac{1}{4 π ε_{0}} \frac{(+ 3 μ C) (+ 6 μ C)}{d^{2}} (2)$
Dividing equ (2) eqn (1), we get
$\frac{F^{'}}{F} = \frac{\frac{1}{4 π ε_{0}} \frac{(+ 3 μ C) (+ 6 μ C)}{d^{2}}}{\frac{1}{4 π ε_{0}} \frac{(+ 6 μ C) (+ 9 μ C)}{d^{2}}}$
$= \frac{(+ 3 μ C) (+ 6 μ C)}{(+ 6 μ C) (+ 9 μ C)} = \frac{1}{3}$ or $F^{'} = \frac{F}{3}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358071 Assertion :
Two charges $q_{1}$ and $q_{2}$ , are placed at separation $r$ . Then, magnitude of force $(F)$ on each charge is given by coulomb law.
Reason :
Now a third charge $q_{3}$ is placed near $q_{1}$ and $q_{2}$ Then force on $q_{1}$ , due to $q_{2}$ , remains $F$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII01:ELECTRIC CHARGES AND FIELDS

358072 When a brass plate is introduced between two charges, the force between the charges

1 Decreases

2 Increases

3 Remains same

4 Becomes zero

Explanation:

Force becomes zero after inserting the slab as $k = \infty$

PHXII01:ELECTRIC CHARGES AND FIELDS

358073 Two small spheres each having the charge $+ Q$ are suspended by insulating threads of length $L$ from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between two suspensions and the tension in each will be

1

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{{(2 L)}^{2}}

2

90^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{L^{2}}

3

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{2 L^{2}}

4

180^{0}, \frac{1}{4 π \in_{0}} \frac{Q^{2}}{L^{2}}

Explanation:

In the absence of gravitational force, the only force acts on the spheres is electrostatic repulsion and so the angle between two suspension becomes $180^{0}$ . So force between the spheres
$F = \frac{1}{4 π \in_{0}} \frac{Q^{2}}{{(2 L)}^{2}}$

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