A copper slab of thickness b = 1.194 mm is inserted at constant speed into a par

Question

A copper slab of thickness b = 1.194 mm is inserted at constant speed into a parallel-plate capacitor with gap d = 8.0 mm, as shown in the figure. The slab Is centered exactly halfway between the plates. If the initial capacitance was C = 4.00E - 11 F, what is the capacitance after the slab is introduced? Tries 0/7 If a charge q = 6.00E-6 C is maintained on the plates as the slab is inserted, what is the ratio of the initial stored energy to the final stored energy? Tries 0/7 How much work Is done by the person inserting the slab?

Explanation / Answer

(a)
Initial Capacitance, A*e0/d = 4.0 * 10^-11 F

The new arrangement is equivalent to having two capacitors connected in series -
C' = A*e0/((d - b)/2)
Ct = C'/2
Ct = A*e0/((d - b))
Ct = (d * 4.0 * 10^-11 ) / (d-b)
Ct = (8 * 4.0 * 10^-11) / (8 - 1.194)
Ct = 4.70 * 10^-11 F

(b)
Initial Stored Energy = 1/2 * Q^2/C = 1/2 * ((6*10^-6)^2 / (4.0 * 10^-11)) = 0.45 J
Final Stored Energy = 1/2 * Q^2/Ct = 1/2 * ((6*10^-6)^2 / (4.7 * 10^-11)) = 0.383 J

Ratio of Initial Stored Energy/ Final Stored Energy = 0.45/0.383
Ratio of Initial Stored Energy/ Final Stored Energy = 1.175

(c)
Work done  = 0.383 - 0.45 J
Work done = - 0.067 J

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