10 A parallelplate capacitor has square plates of side s 250

10. A parallel-plate capacitor has square plates of side s 2.50cm and +0 plate separation d = 2.50mm. The capacitor is charged by a battery to a charge Q=4.00AC, after which the battery is disconnected. A porcelain dielectric (K = 6.5) is then inserted a distance y = 1.00cm into the capacitor as shown in the figure. Hint: Consider the system as two capacitors connected in parallel. Dielectric a) What is the effective capacitance of this capacitor? b) How much work is done in inserting the dielectric?

Solution

before inseting dielectric

capacitance of the capacitor co = eo*A/d = eo*s^2/d = 8.84*10^-12*0.025^2/(2.5*10^-3)

co = 2.21*10^-12 F

A = area of plate = s^2

after the dielctric is inserted

capacitance of upper part c1 = eo*s*(s-y)/d

capacitance of lower part c2 = k*eo*s*y/d

C1 and C2 are parallel

Ceff = C1 + C2 = (eo*s/d)*( s - y + ky)

Ceff = (8.84*10^-12*0.025/(2.5*10^-3))*(0.0025 - 0.01 + (6.5*0.01))

Ceff = 7.07*10^-12 F

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b)

initial energy Ui = (1/2)*Q^2/Co = (1/2)*(4*10^-6)^2/(2.21*10^-12) = 3.62 J

final energy Uf = (1/2)*Q^2/Ceff = (1/2)*(4*10^-6)^2/(7.07*10^-12) = 1.31 J

work = dU = 2.31 J