A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 μC. What is the potential difference between the plates? Express your answer with the appropriate units. Part B If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? Express your answer with the appropriate units. Part C How much work is required to double the separation? Express your answer with the appropriate units (mJ)

Answer: a) 235.9*10^-6 V; b) 471.8 * 10^-6 V; c) 1.08*10^-13 J

Explanation: In order to explain this problem we have to use the expression for a capacitor od two parallel plates, which is given by:

C=Q/V where Q and V are the charge and voltage difference in teh capacitor, respectively.

Then we have;

V=Q/C=920*10^-12 / (3.9*10^-6) = 235.9*10^-6 V

If we double the separation between the plates, the potencial difference will be: (Q is constant)

Vnew=Q/Cnew where Cnew is given by;

Cnew=ε * A / (2*d) = 1/2 * (ε * A/d) = 1/2 * Co; then:

Vnew=2*Q / (Co) = 2*Vo=2*4.24 * 10^3 V=471.8 * 10^-6 V

Finally, the work made oin the capacitor is the difference of the final and initial potential energy so:

Uo=Q^2 / (2Co) =

Uf = Q^2 / (2Cnew) Cnew=Co/2

W=Uf-Uo = Q^2 (1/Co) - Q^2 (1/2*Co) = Q^2 (1/2*Co) = 1.08*10^-13 J