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A parallel-plate capacitor with plates of area 660 cm2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.3 cm farther apart, the voltage between the plates increases by 100 V.

(a) What is the charge Q on the positive plate of the capacitor?

(b) How much does the energy stored in the capacitor increase due to the movement of the plates?
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The area of the parallel plate capacitor, A = 660 cm2
= (660 cm2)(10-4 m2/1 cm2)
= 0.066 m2

The separation between two plates, ?r = 0.3 cm
= (0.3 cm)(10-2 m/1 cm)
= 0.310-2 m

The voltage between the plates , ?V = 100 V

The relation between the charge (Q) on positive plate of the capacitor and the charge density(?) of the plate's is
Q = ?A      ...... (1)

The relation between the charge dansity and the electric field(E) between the plates of the capacitor is
? = ?0E    ...... (2)

Substitute the equation (2) in equation (1), we get
Q = ?0EA ...... (3)

The relation between the electric field between the plates of the capacitor and the potential difference (?V) is
E = ?V/?r     ...... (4)

Substitute the equation (4) in equation (3), we get
Q = ?0A(?V/?r)    ...... (5)

Here, the permitivity of the free space ?0 = 8.8510-12 C2/N.m2

Using equation (5), the charge Q on the positive plate of the capacitor is
Q = (8.8510-12 C2/N.m2)(0.066 m2)(100 V)/(0.310-2 m)
= (17.52310-9 C)(1 nC/10-9 C)
= 17.523 nC

The energy stored in the capacitor is
?U = (1/2)C(?V)2          ...... (6)

The relation between the charge on the positive plate and
the capacitance (C) of the capacitor is
Q = C?V
C = Q/?V      ...... (7)

Substitute the equation (7) in equation (6), we get
?U = (1/2)(Q/?V)(?V)2
= (1/2)(Q)(?V)     ...... (8)

Using equation (8), the energy stored in the capacitor is
?U = (1/2)(17.52310-9 C)(100 V)
= (0.87610-6 J)(1 ?J/10-6 J)
= 0.876 ?J
= 0.88 ?J     (approximtely)

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