A slab of copper of thickness b = 1.167 mm is thrust into a parallel-plate capacitor of C = 7.00×10-11 F of gap d = 7.0 mm, as shown in the figure; it is centered exactly halfway between the plates.If a charge q = 4.00×10-6 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?
Unfortunately the mentioned figure seems to be absent but, strangely, I like the problem so let's see what I can do without the figure. By the way, engineers could do a little for their dull image by simply saying please if they want a favour. A slab of copper has even less charm and excitement than a refrigerator! The capacitance C ∝ A/d where A is the area of the plates and d the distance between them. ∝ is the proportionality sign. The initial stored energy in a capacitor = E = ?.C?.V?? In this case the initial voltage V? on the capacitor = Q/C? = 4×10??/(7×10???) = 57,143 volts. In this case, the initial energy = E? = ?×7×10??? ×(57143)? = 0.1143 J. By insertion of the copper plate the capacitance increases to C? = C?×7/5.833 = 8.4×10??? Farad At constant Q the voltage then reduces to Q/C?= 47,619 volts and the stored energy to E? = ?×8.4×10???×(47,619)? = 0.0952 J Ratio E?/E? = 1.20 Work done by the field on the copper = 0.1143 – 0.0952 = 0.019 J.
