Question

. A parallel-plate capacitor has fixed charges +Q and –Q. The separation of the plates is then doubled. (a) By what factor does the energy stored in the electric field change? (b) How much work must be done if the separation of the plates is doubled from d to 2d? The area of each plate is A. Now consider a cylindrical capacitor with inner and outer radii a and b, respectively. (c) Suppose the outer radius b of a cylindrical capacitor is doubled, but the charge is kept constant. By what factor would the stored energy change? Where would the energy come from? (d) Repeat (c), assuming the voltage remains constant.

Answer 1

The energy stored in the electric field of a parallel-plate capacitor decreases by a factor of 2 when the separation of the plates is doubled. The work done in doubling the separation is equal to -(1/4)CV^2. When the outer radius of a cylindrical capacitor is doubled while keeping the charge constant, the energy stored in the electric field doubles, and the energy comes from the work done in stretching the electric field lines.

Step-by-step explanation:

(a) The energy stored in the electric field of a parallel-plate capacitor is given by the formula:

E = (1/2)CV^2

Since the capacitance C is inversely proportional to the separation of the plates d, doubling the separation will result in half the capacitance. Therefore, the energy stored in the electric field will decrease by a factor of 2.

(b) To calculate the work done in doubling the separation from d to 2d, we can use the formula:

W = ΔU

Since the energy stored in the electric field is given by E = (1/2)CV^2, the change in energy is ΔU = (1/2)(C1 - C2)V^2, where C1 and C2 are the initial and final capacitances respectively. Doubling the separation will result in half the capacitance, so the change in energy is given by:

ΔU = (1/2)(C/2 - C)V^2 = - (1/4)CV^2

Therefore, the work done must be equal to the change in energy, which is - (1/4)CV^2.

(c) When the outer radius b of a cylindrical capacitor is doubled, but the charge is kept constant, the capacitance will also double. The energy stored in the electric field is given by E = (1/2)CV^2, so doubling the capacitance will result in double the energy stored. The energy comes from the work done in stretching the electric field lines from the inner to the outer plates.

(d) If the voltage remains constant while the outer radius b of a cylindrical capacitor is doubled, the capacitance will also double. Since the energy stored in the electric field is given by E = (1/2)CV^2, doubling the capacitance will result in double the energy stored. The energy still comes from the work done in stretching the electric field lines from the inner to the outer plates.