Mastering Physics Solutions: Energy in Capacitors and Electric Fields

Energy in Capacitors and Electric Fields

Part A = U = Q2 / (2C)
Part B = U = (CV2) / 2
Part C = U/U0 = 0.5
Part D = U/U0 = 2
Part E = U = (ε0AV2) / 2d
Part F = U = (ε0AdE2) / 2
Part G = u = (ε0E2) / 2

Solutions Below:

First, consider a capacitor of capacitance C that has a charge Q and potential difference V.

Part A

Find the energy U of the capacitor in terms of C and Q by using the definition of capacitance and the formula for the energy in a capacitor.
Express your answer in terms of C and Q.

Start with the formula U = QV/2, which should be given in the problem (if not, you have it now). Recall the definition of capacitance: C = Q/V. You can rearrange that as V = Q/C. So:

U = QV / 2
U = (Q * (Q/C)) / 2

U = Q2 / (2C)

Part B

Find the energy U of the capacitor in terms of C and V by using the definition of capacitance and the formula for the energy in a capacitor.
Express your answer in terms of C and V.

Again, start with U = QV/2. The definition of capacitance is C = Q/V, so rearrange as Q = CV:

U = QV/2
U = ((CV) * V) / 2

(CV2) / 2

Part C

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0. The capacitor remains connected to the battery while the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U0.

When the plates are pulled apart, the voltage remains constant (since the capacitor remains connected to the battery, but the capacitance is halved. So use the formula from Part B:

Initial: U = (CV2) / 2
Final: ((C / 2)V2) / 2

Divide final (U) by initial (U0). V2 cancels:

U/U0 = [(C / 2) / 2] / [C / 2]
U/U0 = (C/4) / (C/2)

U/U0 = 0.5

Part D

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0. The capacitor is then disconnected from the battery and the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U0.

This time, the voltage doesn’t remain constant. So we can’t use the formula that has the voltage term in it. Use the one from Part A instead. Remember the capacitance will be halved when the distance doubles:

Initial: U = Q2 / (2C)
Final: U = Q2 / (2(C / 2))

Divide final (U) by initial (U0). Q2 cancels:

U/U0 = [2C] / [(2(C / 2))]
U/U0 = 2C / C

U/U0 = 2

Part E

A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem introduction to obtain the formula for the energy U of the capacitor.
Express your answer in terms of A, d, V, and appropriate constants.

C = ε0A/d, which should have been given in the problem (if not, you have it now). Start with the formula from Part B (you can’t use Part A because it has a “Q” term):

U = (CV2) / 2

Substitute in for C:
U = (ε0A/d) * V2 / 2

U = (ε0AV2) / 2d

Part F

A parallel-plate capacitor has area A and plate separation d, and it is charged so that the electric field inside is E. Use the formulas from the problem introduction to find the energy U of the capacitor.
Express your answer in terms of A, d, E, and appropriate constants.

Mastering Physics says to use the formulas from the problem introduction, but it doesn’t actually give you one of the ones you need. Go figure. Anyway, start with the formula from Part E:

U = (ε0AV2) / 2d

And recall that V = Ed, so V2 = E2 * d2.

Substitute in for V:

U = (ε0A * (E2 * d2)) / 2d

Simplify to get the final answer:

U = (ε0AdE2) / 2

Part G

Find the energy density u of the electric field in a parallel-plate capacitor. The magnitude of the electric field inside the capacitor is E.
Express your answer in terms of E and appropriate constants.

The formula for energy density is u = U / V (the V here stands for volume, not voltage. Start with the formula from Part F:

U = (ε0AdE2) / 2

The Ad term is equivalent to the volume of a symmetrical box (A is the area of a base, d is the distance between the bases). So divide U by the volume (Ad):

U = ((ε0AdE2) / 2) / (Ad)

Simplify to get the final answer:

u = (ε0E2) / 2

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2 Responses to Mastering Physics Solutions: Energy in Capacitors and Electric Fields

  1. Joss Mon says:

    great job!!!! :)

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