(8c25p74) A slab of copper of thickness b = 1.139 mm is thrust into a parallel-p

Question

(8c25p74) A slab of copper of thickness b = 1.139 mm is thrust into a parallel-plate capacitor of C = 1.00×10-11 F of gap d = 9.0 mm, as shown in the figure; it is centered exactly halfway between the plates. What is the capacitance after the slab is introduced?

If a charge q = 6.00×10-6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted?

How much work is done on the slab as it is inserted?

Is the slab pulled in or must it be pushed in?

Explanation / Answer

a)

C = {(*A)/0.009} = 1.00*10¹¹ F ---------------- (1)

The new combined capacitance, C1 = [(1/2)*{(*A)/(0.5*(0.009-0.001139))}] or
C1 = [ (*A)/(0.007861)] ------------------------- (2)
(2)/(1) gives
C1 = (0.009/0.007861)*C or
C1 = 1.14*C = 1.14*10¹¹ F

b)

Stored energy before, E1 = [Q²/(2*C)] = (36*10¹²)/(2*1011 ) =1.8 J
Stored energy after, E2 = [Q²/(2*C1)] = (36*10¹²)/(2.28*1011 ) =1.5 J
Worked done is negative = -0.3 J

c) since the workdone is negative, it may be pulled.

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