Question

Two charges $3\times 10^{-8}C$ and $-2\times 10^{-8}C$ located $15cm$ apart. At what point on the line joining the two charges is the electric potential zero?

• A. 9 cm
• B. 45 cm
• C. 18 cm
• D. Both (a) and (b)

Answer 1

Answer: (D)

$\frac{1}{4\pi {\epsilon }_0}\left[\frac{3\times 10^{-8}}{x\times 10^{-2}}-\frac{2\times 10^{-8}}{(15-x)\times 10^{-2}}\right]=0$
where, $x$ in $cm$, i.e., $\frac{3}{x}-\frac{2}{15-x}=0$
Which gives $x=9cm$
If $x$ lies on the extended line $OA$, the required condition is:
$\frac{3}{x}-\frac{2}{x-15}=0$
Which gives $x=45cm$
Thus, electric potential is zero at $9cm$ and $45cm$ away from the positive on the side of the negative charge.