Pictured at right is a parallel plate capacitor C, made up of two square plates
ID: 2236770 • Letter: P
Question
Pictured at right is a parallel plate capacitor C, made up of two square plates of side length L separated by a distance d, where d << L. A battery of potential difference Vb is attached along with a switch and a resistor, R, to the capacitor. The switch is initially set to position 1, which charges the capacitor until it is fully charged.
a) If L = 5m and d = 0.0001 m and ?e = 310 and Vb = 12 V, determine the change in the electric potential energy stored in the capacitor from before to after inserting the dielectric. Do we have to push the dielectric in, or is it pulled in? (Remember, things
Explanation / Answer
a)
C = epsilon0 L^2/d = 8.85e-12*5*5/0.001 = 0.00000022125 F
C' = (epsilon0 L^2/d) * (4/3 + 4 ke) = 8.85e-12*5*5/0.001*(4/3+4*310) = 0.000274645 F
Q = CV = 0.00000022125 * 12 = 0.000002655 C
the change in the electric potential energy:
P - P' = 0.5 Q^2/C - 0.5 Q^2/C'
= 0.5*0.000002655*0.000002655/0.00000022125 - 0.5*0.000002655*0.000002655/ 0.000274645
= 1.5917e-5 J
average force:
F = (P - P')/(L/2) = 1.5917e-5/(5/2) = 6.37 * 10^-6 N
b)
Q/Qo = e^(-t/RC')
0.5 = e^(-t/(0.000274645*1e3))
t = 0.190 seconds