Question 920: Electrostats & Capacitance

Question

Two point charges A and B, having charges +Q and – Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes :

Accepted Answer

Initially F = $$ - {{K{Q^2}} \over {{r^2}}}$$
If 25% of charge of A transferred to B then after the
transformation of charge.
q
A
= $$Q - {Q \over 4}$$ = $${{3Q} \over 4}$$
and q
b
= -Q + $${Q \over 4}$$ = $$ - {{3Q} \over 4}$$
New force, F
1
= $${{K\left( {{{3Q} \over 4}} \right)\left( { - {{3Q} \over 4}} \right)} \over {{r^2}}} $$
$$=  - {9 \over {16}}{{K{Q^2}} \over {{r^2}}}$$
= $${{9F} \over {16}}$$

Answer

F

Answer

$${{16F} \over 9}$$

Answer

$${{9F} \over {16}}$$

Answer

$${{4F} \over 3}$$

Electrostats & Capacitance

Two point charges A and B, having charges +Q and – Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes :

Choose the correct answer: