A parallel-plate capacitor is charged by being connected To a battery and is then disconnected from the battery. The separation between the plates is then doubled. How does the electric Field change? The potential difference? The total energy? Explain.

The difference of electrical potential between two points. Amount of work done in moving a unit charge from one point.

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

The total energy stored in the capacitor depends on the charge Q and the potential difference V between the two plates as shown in the equation 24.11 in the next form

As we discussed the charge is constant while the potential difference is doubled when the distance is doubled and as shown by equation (2) the energy stored is directly proportional to the potential difference V. So we could conclude that the total energy stored U is doubled when the distance d is doubled.

Therefore , The total energy stored U is doubled