First slide

Electrostatic force

Question

There are two equally charged identical metal spheres A and B repel each other with a force 3 $\times$ 10^-5 N. Another identical uncharged sphere C is touched with A and then placed at the mid-point between A and B. Net force on C is

Moderate

A

$1 \times 10^{- 5} N$
B

$2 \times 10^{- 5} N$
C

$1.5 \times 10^{- 5} N$
D

$3 \times 10^{- 5} N$

Solution

If same charges on spheres A and Bare q, then force,
$F = \frac{1}{4 {πε}_{0}} \frac{q^{2}}{r^{2}} = 3 \times 10^{- 5} N$
Charge on A and C after touching,
$q_{A}^{'} = q_{C} = \frac{q}{2}$
Net force on C,
$\begin{matrix} F = F_{A} + F_{B} \\ ∴ F = |\frac{1}{4 {πε}_{0}} \frac{(q / 2) (q / 2)}{(r / 2)^{2}} - \frac{1}{4 {πε}_{0}} \frac{(q / 2) \times q}{(r / 2)^{2}}| \\ = |\frac{1}{4 {πε}_{0} r} \frac{q^{2}}{r^{2}} - 2 \frac{1}{4 {πε}_{0}} \frac{q^{2}}{r^{2}}| \\ = |- (\frac{1}{4 {πε}_{0}} \cdot \frac{q}{r^{2}})| = 3 \times 10^{- 5} N \end{matrix}$

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