The charge and mass of two particles are +Q, M and -q, m respectively. The particles, separated by a distance L, are released from rest in a uniform electric field E. The electric field is parallel to the 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: $(K = \frac{1}{4 π \in_{0}})$

\sqrt{\frac{(M + m) K Q q}{E (q M + Q m)}}

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\sqrt{\frac{(M + m) K Q q}{E (q m + Q M)}}

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\sqrt{\frac{m M K Q q}{E (q M + Q m)}}

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\sqrt{\frac{m M K Q q}{E (Q M + q m)}}

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Solution

The correct option is A $\sqrt{\frac{(M + m) K Q q}{E (q M + Q m)}}$
In order for the distance between them to be a constant, the relative motion between them has to be zero i.e. $a_{r e l a t i v e} = 0$

$M a_{M} = Q E - \frac{k Q q}{L^{2}}$

$m a_{m} = - q E + \frac{k Q q}{L^{2}}$

$a_{r e l a t i v e} = 0 \Rightarrow a_{m} = a_{M}$

$\frac{Q E}{M} - \frac{k Q q}{(L^{2} M)} = \frac{- q E}{m} + \frac{k Q q}{(L^{2} m)}$

$\frac{E}{M m} \times (Q m + q M) = \frac{k Q q}{L^{2}} \times \frac{(M + m)}{M m}$

$L = \sqrt{\frac{k Q q (M + m)}{E (Q m + q M)}}$

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The correct option is A √(M+m)KQqE(qM+Qm)In order for the distance between them to be a constant, the relative motion between them has to be zero i.e. arelative=0MaM=QE−kQqL2mam=−qE+kQqL2arelative=0⇒am=aMQEM−kQq(L2M)=−qEm+kQq(L2m)EMm×(Qm+qM)=kQqL2×(M+m)MmL=√kQq(M+m)E(Qm+qM)