Question

An isolated parallel-plate capacitor (not connected to a battery) has a charge of Q = 4.8×10−5C. The separation between the plates initially is d = 1.2 mm, and for this separation the capacitance is 3.1×10−11F. Calculate the work that must be done to pull the plates apart until their separation becomes 9.1 mm, if the charge on the plates remains constant. The capacitor plates are in a vacuum.

Answer 1

Given

capacitor with charge Q = 4.8*10^-5 C

separation is d1 = 1.2mm , C1 = 3.1*10^-11F

work should be done for d2 = 9.1 mm, U2 =?

we know that the capacitance of a parllel plate capacitor is  

C = epsilon_0*A/d

wehre epsilon_0 is permitivity of free space (vacuum = 8.854*10^-12 C^2/N.m^2)

A is are of the plates (here constant )

d is separation of plates

that is C is inversly proportional to d

C2/C1 = d1/d2

C2 = d1*C1/d2

substituting the values

C2 = (1.2*10^-3*3.1*10^-11)/(9.1*10^-3) F

C2 = 4.0879120*10^-12 F or = 4.1 pF

we know that the work done = potential energy  

so potential energy stored in the capacitor  

W = q^2/2C

W = q^2/2(1/C2-1/c1)

W = (4.8*10^-5)^2*0.5 (1/(4.1*10^-12)-1/(3.1*10^-11) ) J

W = 243.81 J

work that must be done to pull the plates apart until their separation becomes 9.1 mm, if the charge on the plates remains constant is 243,81 J