A parallel-plate capacitor has capacitance 6.00 mu F. (a) How much energy is sto
ID: 1652051 • Letter: A
Question
A parallel-plate capacitor has capacitance 6.00 mu F. (a) How much energy is stored in the capacitor if it is connected to a 9.00 V battery? (b) If the distance between the charged plates doubled, what is the energy stored? (c) If the battery is now disconnected and the distance between the charged plates doubled, what is the energy stored? (d) The battery is subsequently reattached to the capacitor, but the plate separation remains as in part How much energy is stored? (e) The plates are then moved back to the original distance between them. What is the potential difference across the capacitor?Explanation / Answer
4)Given, C = 6 uF = 6 x 10^-6 F
a)We know that stored energy in capacitor is given by:
E = 1/2 C V^2
E = 0.5 x 6 x 10^-6 x 9^2 = 2.43 x 10^-4 J
Hence, E = 2.43 x 10^-4 J
b)We know that, capacitance is related to the distance between plates as:
C = epsilon0 A/d
C' = epsilon0 A/2d
C/C' = 2d/d = 2 => C' = C/2
C' = 6 x 10^-6/2 = 3 x 10^-6 F
E' = 1/2 C' V^2 = 0.5 x 3 x 10^-6 x 9^2 = 1.22 x 10^-4 J
Hence, E' = 1.22 x 10^-4 J
c)Charge is conserved.
Q = 6 x 9 = 54 uC ; C' = 3 x 10^-6 F
E = 1/2 C V^2 = 1/2 Q^2/C = 0.5 x (54 x 10^-6)^2/3 x 10^-6 = 4.86 x 10^-4 J
Hence, E = 4.86 x 10^-4 J
d)C' = 3 x 10^-6 F ; V = 9 V
E = 1/2 C V^2 = 0.5 x 3 x 10^-6 x 9^2 = 1.22 x 10^-4 J
Hence, E = 1.22 x 10^-4 J
e)charge is conserved.
Q = CV => V = Q/C = 54 uC/6 = 9 V