First slide

Electrostatic force

Question

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases 4.5 times in comparison with the initial value. The ratio of the initial charges of the balls is

Moderate

A

2
B

3
C

4
D

6

Solution

Suppose the balls have charges Q₁ and Q₂ respectively.
Initially,
Finally,
$F^{'} = \frac{k {(\frac{Q_{1} + Q_{2}}{2})}^{2}}{{(\frac{r}{2})}^{2}} = \frac{k {(Q_{1} + Q_{2})}^{2}}{r^{2}}$
$It is given that F' = 4.5 F \Rightarrow \frac{k {(Q_{1} + Q_{2})}^{2}}{r^{2}} = 4.5 k \cdot \frac{Q_{1} Q_{2}}{r^{2}}$
$\Rightarrow {(Q_{1} + Q_{2})}^{2} = 4.5 Q_{1} Q_{2} . Let say \frac{Q_{1}}{Q_{2}} =x \Rightarrow {(x + 1)}^{2} = 4.5 x \Rightarrow 2 x^{2} - 5 x + 2 = 0 \Rightarrow x = 2 o r \frac{1}{2} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{2}{1} .$

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