Question

The distance between charges $5\times 10^{-11}C$ and $-2.7\times 10^{-11}C$ is $0.2m$. The distance at which is third charge should be placed from $4e$ in order that it will not experience any force along the line joining the two charges is

• A. 0.44 m
• B. 0.65 m
• C. 0.556m
• D. 0.350m

Answer 1

Answer: (C)

From the question $F_1=F_2$
$5\times 10^{-11}C-2.7\times 10^{-11}C$

$\Rightarrow \frac{1}{4\pi {\epsilon }_0}\times \frac{5\times 10^{-11}\times q}{(0.2+x{)}^2}$
$=\frac{1}{4\pi {\epsilon }_0}\times \frac{2.7\times 10^{-11}\times q}{x^2}$
$\Rightarrow x=0.556m$ from IInd charge.