Posted on May 10, 2010.
What is the charge Q on the positive plate of capacitor? A parallel plate capacitor with plates of area 420 cm2, is charged to a potential difference V, then is disconnected from the power source. When the plates are moved further away by 0.3 cm, the voltage between the plates increases by 100 V.
(A) What is the charge Q on the positive plate of capacitor?
(B) What is the energy stored in the increase in capacity due to the movement of plates?
ε ae = 8.854e-12 C ² N ¹ a a ² m
S = 0.042 sq.m.
Δd = 0.003 m
ΔU = 100 V
d = initial distance between the plates
U = initial tension between the plates
Q = load on the positive plate
(A) C ae ε = S / D is the initial capacity of parallel plate capacitor
C1 ae ε = S / (d + Δd) is the capacity after moving plates
U = Q / C
U + ΔU = Q/C1 is the voltage after the removal of the plate ==>
Qd / (ε ae S) + ΔU = Q (d + Δd) / (ε ae S) ==>
ΔU = Q * Δd / (ε ae S)
Q = S * ae ε ΔU / Δd
Q = 8.854e-12 * 0.042 * 100/0.003 = 12.4e-9 C
(B) W = initial energy 0.5QU
W1 = 0.5Q (U + ΔU)
ΔW = W1-W = ΔU = 0.5Q * 0.5 * 12.2E-9 * 100
ΔW = 0.62e-6 J
