Two particles A & B having charges $+ q$ and $- 2 q$ . What happens to Electrostatic force between them if the distance between the A & B is halved?

It decreases by a factor of

4

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It decreases by a factor of

2

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It remains the same

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It increases by a factor of

2

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It increases by a factor of

4

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Solution

The correct option is C It increases by a factor of $4$
The electrostatic force between two charges is $F = \frac{k (q) (- 2 q)}{r^{2}}$
When distance r is halved, the force $F^{'} = \frac{k (q) (- 2 q)}{(r / 2)^{2}} = 4 F$

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Two particles A & B having charges +q and −2q. What happens to Electrostatic force between them if the distance between the A & B is halved?

The correct option is C It increases by a factor of 4The electrostatic force between two charges is F=k(q)(−2q)r2When distance r is halved, the force F′=k(q)(−2q)(r/2)2=4F

Two particles A & B having charges $+ q$ and $- 2 q$ . What happens to Electrostatic force between them if the distance between the A & B is halved?

The correct option is C It increases by a factor of $4$
The electrostatic force between two charges is $F = \frac{k (q) (- 2 q)}{r^{2}}$
When distance r is halved, the force $F^{'} = \frac{k (q) (- 2 q)}{(r / 2)^{2}} = 4 F$