A parallel-plate capacitor is charged by being connected To a battery and is kept connected to the battery. The separation Between the plates is then doubled. How does the electric field Change? The charge on the plates? The total energy? Explain.

Electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

Given

The parallel plates are kept connected to the battery. This means the applied voltage is fixed. Also, the distance d between the two plates is doubled.

Required

The change in the electric field E, the charge Q and the total energy U

Explanation

The electric field depends on the separated distance between the two plates and it is given by

As shown by equation (1), the electric field is inversely proportional to the separated distance d and the potential difference is constant. Hence, the electric field is halved when the distance d is doubled.

Therefore The electric field E is halved when the distance d is doubled

The distributed charge over the plates is related to the potential difference and the capacitance of the capacitor in the next form

But the capacitance is related to the distance between the two plates as shown by equation 24.2

Let us use this experience for C into equation (2), we will find that

As shown by this equation,

The charge is inversely proportional to the distance d. As the distance doubled the charge is halved where the area and the potential t are constant. Actually, this occurs when the potential is held constant as the capacitor is held connected to the battery.

Therefore The charge Q is halved when the distance d is doubled.

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

As we discussed the charge is halved when the distance is doubled and as shown by equation (3) the energy stored is directly proportional to the charge Q. So we could conclude that the total energy stored U is halved when the distance d is doubled.

Therefore The total energy stored U us halved when the distance d is doubled.

Result

The electric field E is halved when the distance d is doubled

The charge Q is halved when the distance d is doubled.

The total energy stored U us halved when the distance d is doubled.