A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
(i) charge stored by the capacitor.
(ii) Field strength between the plates.
(iii) Energy stored by the capacitor.
Justify your answer in each case.
Solution
(i)The charge stored on the capacitor does not change because of the law of conservation of charges.
(ii) The field strength between the plates is
`E=sigma/epsilon_0=Q/(Aepsilon_0)`
Hence, we see that the field strength is independent of the distance between the plates. So, the field strength also remains the same.
(iii)The energy stored in the capacitor is
`W=Q^2/(2C)`
Now, when the distance between the plates is doubled, the capacitance becomes half. Hence, the energy stored will also double.
