Question

how will the capacitance be affected if separation between it plate is doubled and the size of the plate is reduced to half?

Answer 1

C = €oA/d

Now its 2d in place of d and A/2 in place of A

Therefore C' = €o (A/2)/(2d)

This gives C' = €oA/(4d)

=> C' = (€oA/d)/ 4

=> C' = C/ 4

Hence the Capacitance will be reduced by a factor of 4 i.e the new capacitance will be one fourth of its original value.