A) Each plate of a parallel-plate capacitor is a square of side 4.97 cm and the

Question

A) Each plate of a parallel-plate capacitor is a square of side 4.97 cm and the plates are separated by 0.423 mm. The capacitor is charged and stores 7.93 nJ of energy. Find the electric field strength inside the capacitor.

= ____________ N/C

B) You have two capacitors, one with capacitance 16.9 F and the other of unknown capacitance. You connect the two capacitors in series with a voltage difference of 383-V applied across the capacitors. You discover that, as a result, the unknown capacitor has a charge of 1.11 mC. Find its capacitance.

= ___________ F

Explanation / Answer

Part A)

The electric Field in a Capacitor is found by

V = Ed

so E = V/d

The voltage can be found from the Energy, where Energy = .5CV2

V = (2(Energy)/C)

C is the capacitance found by C = oA/d

So, with all that substitution, we find that E = 1/d ((2(Energy)d/oA)

E = (1/4.23 X 10-4)([(2)(7.93 X 10-9)(4.23 X 10-4)/(8.85 X 10-12)(2.47 X 10-3)])

E = 4.14 X 104 N/C

Part B)

From Q = CV, we can find the total Capacitance

Since the capacitors are in series, the value for Q stays constant, so that is the total Q of the circuit

(1.11 X 10-3) = (C)(383)

C = 2.90 X 10-6   which is 2.9 F

That total C for capacitors in series is found by

1/CEQ = 1/C1 + 1/C2

1/2.9 = 1/16.9 + 1/C2

C2 = 3.5 F

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