Two point charges having charge +Q, –q and mass M, m respectively are separated by a distance L. They are released from rest in a uniform electric field E. The electric field is parallel to line joining both the charges and is directed from negative to positive charge. For the separation between particles to remain constant, the value of L is
`(K=(1)/(4piin_(0)))`

Options

A. `sqrt(((M+m)KQq)/(E(qM+Qm)))`
B. `sqrt(((M+m)KQq)/(E(qm+QM)))`
C. `sqrt((mmKQq)/(E(qM+Qm)))`
D. `sqrt((mmKQq)/(E(QM+qm)))`

Answer

A

Explanation

In order to maintain constant separation the particles must have the same acceleration Assuming the system of both charges to acceleration towards left.Applying Newton's second law to the left particle we get

`QE-(KQq)/(L^(2))=Ma.......(1)`
Under given condition the acceleration of both charges should be and should also be equal to acceleration of centre of mass of both the charges.
`a=(F_("net"))/("total mass")=((Q-q)E)/(m+M).....(2)`
`L=sqrt((M+m)KQq)/(E(qM+Qm))`