Question

The distance between charges $5 \times 10^{-11} C$ and $-2.7 \times 10^{-11} C$ is $0.2\, m$. The distance at which is third charge should be placed from $4\, e$ 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 \varepsilon_{0}} \times \frac{5 \times 10^{-11} \times q}{(0.2+x)^{2}} $
$=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{2.7 \times 10^{-11} \times q}{x^{2}}$
$\Rightarrow x=0.556\, m$ from IInd charge.