Question

Two charges $12\mu \text{C}$ and $-6\mu \text{C}$ are separated by a distance of $20\text{cm}$. The net electric potential is zero at a point between the charges and on the line joining them will be

Answer 1

Let, $q_1=-6\mu \text{C}{q}_2=12\mu \text{C}$

Distance from the smaller charge(magnitude) is,

$|q_1|=6\mu \text{C},|q_2|=12\mu \text{C}$

$\frac{k|q_1|}{x}=\frac{k|q_2|}{20-x}$

$x=\frac{20}{\frac{q_2}{q_1}+1}$

$x=\frac{20}{\frac{12}{6}+1}=\frac{20}{3}$

$=6.7\text{cm}\text{from}-6\mu \text{C}$

From $12\mu \text{C},$ it is

$d=r-x=20-6.7=13.3\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (b) is the correct answer.