Question

Two charges $3 \times 10^{-8} C$ and $-2 \times 10^{-8} C$ located $15\, cm$ 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)

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