Energy in Capacitors and Electric Fields

Part A = U = Q^{2} / (2C)

Part B = U = (CV^{2}) / 2

Part C = U/U_{0} = 0.5

Part D = U/U_{0} = 2

Part E = U = (ε_{0}AV^{2}) / 2d

Part F = U = (ε_{0}AdE^{2}) / 2

Part G = u = (ε_{0}E^{2}) / 2

**Solutions Below:**

Part A

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 = Q^{2} / (2C)

Part B

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

(CV^{2}) / 2

Part C

_{0}. 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/U

_{0}.

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 = (CV^{2}) / 2

Final: ((C / 2)V^{2}) / 2

Divide final (U) by initial (U_{0}). V^{2} cancels:

U/U_{0} = [(C / 2) / 2] / [C / 2]

U/U_{0} = (C/4) / (C/2)

U/U_{0} = 0.5

Part D

_{0}. 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/U

_{0}.

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 = Q^{2} / (2C)

Final: U = Q^{2} / (2(C / 2))

Divide final (U) by initial (U_{0}). Q^{2} cancels:

U/U_{0} = [2C] / [(2(C / 2))]

U/U_{0} = 2C / C

U/U_{0} = 2

Express your answer in terms of A, d, V, and appropriate constants.

C = ε_{0}A/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 = (CV^{2}) / 2

Substitute in for C:

U = (ε_{0}A/d) * V^{2} / 2

U = (ε_{0}AV^{2}) / 2d

Part F

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 = (ε_{0}AV^{2}) / 2d

And recall that V = Ed, so V^{2} = E^{2} * d^{2}.

Substitute in for V:

U = (ε_{0}A * (E^{2} * d^{2})) / 2d

Simplify to get the final answer:

U = (ε_{0}AdE^{2}) / 2

Part G

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 = (ε_{0}AdE^{2}) / 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 = ((ε_{0}AdE^{2}) / 2) / (Ad)

Simplify to get the final answer:

u = (ε_{0}E^{2}) / 2

great job!!!!

Glad to help!