Question

Three charges $-q,Q,-q$ are placed at equal distance on a straight line. If the total potential energy of the system is zero, then $Q:q$ is:

• A. $1:4$
• B. $-1:4$
• C. $1:2$
• D. $2:1$

Answer 1

Answer: (A)

$1:4$

We know potential energy is a scalar quantity
Now total P.E. $=\frac{k(-q)\left(Q\right)}{x}+\frac{k(-q)\left(Q\right))}{x}+\frac{k(-q)(-q)}{2x}$

given P.E. total $=0$

$0=\frac{-2kqQ}{x}+\frac{k{q}^2}{2x}$

$0=-4Q+q$
$Q:q=1:4$