The charge on the capacitor will remain constant.
and Capacitance will vary as
hence capacitance will decrease by a factoor of 2.
hence, Potential between the plates will increase by a factor of 2.
A parallel-plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected, and the plates pulled apart until their separation is 2d. Derive expression in terms of A, d, and V for:
(a) the new potential difference
(b) the initial and final stored energies, and
(c) the work required to separate the plates
How to do it?