Question

Two charges $4e$ and +e are x distance apart at what distance charge q must be placed from charge +e so that it is in equilibrium?

• A. $\frac{x}{2}$
• B. $\frac{2x}{3}$
• C. $\frac{x}{3}$
• D. None of these

Answer 1

The correct option is C $\frac{x}{3}$

Forces on q due to e$=\frac{hqe}{r^2}$

Forces on q due to $+e=\frac{k4ϵq}{(x-r{)}^2}$

The forces must be equal

$\frac{kqe}{r^2}=\frac{k4ϵq}{(x-r{)}^2}\frac{1}{r^2}=\frac{4}{(x-r{)}^2}3r^2+3xr-xr-x^2=03r(r+x)-x(r+x)=0(3r-x)(r+x)=0r=-xorr=x/3\therefore r=x/3$