Given
The capacitor is connected to the battery then disconnected after charging. This means the applied voltage is not constant and could be changed. Also, the distance d between the two plates is doubled.
Required
The change in the electric field E, the potential difference V and the total energy U
Explanatio
The electric field depends on the separated distance between the two plates and it is given by
But in this case, the potential difference will be changed as d changes. Also, after disconnection; the charge on the plates kept constant and in this case, we could get the electric field in the next form
As the area and thw charge are constant therefore the electric field between both plates is held constant
Step 3: Determine the chage in potential differnce
As shown by equation (1),the potential difference is directly proportional to the distance d.And as the electric field is constant and the distance d is doubled,
Hence the potential difference is, doubled