A 16microF capacitor is charged to 3.2×10^-4C and the charging battery is disconnected.What will be the work required in doubling the distance between the plates
Answers
Answered by
11
Given:
A 16microF capacitor is charged to 3.2×10^-4C and the charging battery is disconnected.
To find:
What will be the work required in doubling the distance between the plates
Solution:
From given, we have,
A 16microF capacitor is charged to 3.2×10^-4C and the charging battery is disconnected.
V = q/C
V = 3.2 × 10^{-4} / 16 × 10^{-6}
∴ V = 20 V
W = F × d
as F = qE
W = q × E × d
as E = V/2d
W = q × (V/2d) × d
W = q × V/2
W = 3.2 × 10^{-4} × 20/2
W = 3.2 × 10^{-4} × 10
W = 3.2 × 10^{-3} J
The work required in doubling the distance between the plates is 3.2 × 10^{-3} J
Similar questions