Two charges particle `(M,+Q)` and `(m,-q)` are placed in a gravity free space where a uniform electric field `E` exist. After the particles are released, they stay at a constant distance from each other. What is the distance? Neglect Gravitational interaction. `k` is electrostatic constant.

Options

A. `sqrt((kqQ)/(E(q+Q))`
B. `sqrt(((M+m)kQq)/(E(qm+QM))`
C. `sqrt(((M+m)kQq)/(E(qM+Qm))`
D. not possible

Answer

c

Explanation

On releasing the particles, if their acceleration is same , they will stay at a constant distace.

`QE-F= Ma,F-qE=ma`
where, `F=(kqQ)/(L^(2))`
From above: `(QE-F)/(F-qE)=(M)/(m)`
`implies (KqQ)/(L^(2))=((mQ_qM)E)/(m+M)implies L= sqrt(((M+m)kqQ)/(E(qM+mQ)))`