Question

Two charges of magnitude $4 \times 10^{-3} C$ and $6 \times 10^{-8} C$ at $A$ and $B$ are $50\, cm$ apart. The electric potential is zero at a point along $A B$, where distance from $A$ is given by

• A. 40 cm
• B. 20 cm
• C. 10 cm
• D. 30 cm

Answer 1

Answer: (B)

Suppose, potential at point $P$ is zero. $V_{1}=V_{2}$
$K \times \frac{4 \times 10^{-8}}{r_{1}}=K \times \frac{6 \times 10^{-8}}{r_{2}}$
$\frac{4 \times 10^{-8}}{x}=\frac{6 \times 10^{-8}}{(50-x)}$
$4(50-x)=6 x $
$200-4 x=6 x $
$\Rightarrow 10 x=200 x=20\, cm$