Question

please provide solution
i will select brainliest answer!!!!

Answer 1

Answer:

this is a answer but don't mark it as brillianist because I just copied it from the web.

Answer 2

Answer:

Work done will be equal to change in electric potential energy

Initial potential energy

when charge Q is on C

distance AC = BC = L ( mid point )

$intial = \frac{kQq}{L} + \frac{kQ( - q)}{L} = 0$

Final potential energy

when Q is on point D

Distance , AD = 3L and BD = L

$final = \frac{kQq}{3L} + \frac{kQ( - q)}{L} = \frac{ - 2kQq}{3L}$

So difference

$= final - intial \\ \\ = \frac{ - 2kQq}{3L} - 0 \\ \\ = \frac{ - 2Qq}{4\pi \epsilon _{0}.3L} \\ \\ \green{ = \frac{ - Qq}{6\pi \epsilon _{0}L}}$

Option B