Question

A charge of $10\mu c$ and $-10\mu c$ are placed 1cm apart. A third charge of is to be placed on the line passing through the two charges such that it is at equilibrium. Find the position of the third point. How does your answer change if the magnitude of the third charge is double?

Answer 1

Let the third charge be $+Q$

Let the charges be in the equilibrium

Then the net force on $+Q$ will be zero

$\frac{K\times 10\times {10}^{-6}\times Q}{x^2}=\frac{K\times 10\times {10}^{-6}\times Q}{{(0.01+x)}^2}$

$+{(0.01+x)}^2=x^2$

${10}^{-4}+x^2+0.02x=+x^2$

${10}^{-4}=+0.02x$

$x=\frac{{10}^{-4}}{2\times {10}^{-2}}$

$x=0.5\times {10}^{-2}$

$x=0.5cm$