Problem 7: Two capacitors of 2μf and 8μf are joined in series and a potential difference of 300V is applied. Find the charge and potential difference for each capacitor.

Solution:

Step: 1 Overview

See the overview of problem 8.

Step: 2 Calculation

Since capacitors are in series, so their equivalent capacitance will be:

1/C= 1/2 + 1/8
1/C = 5/8
C = 8/5
C = 1.6μf

Total charge will be:

q = CV
q = (1.6x10^-6)(300)
q = 4.8x10^-4C

Due to series combination, q₁=q₂=q=4.8x10^-4C.

Voltage across 2μf capacitor:

V₁=q₁/C₁
V₁=(4.8x10^-4)/(2x10^-6)
V₁=240V

Voltage across 8μf capacitor:

V₂=q₂/C₂
V₂=(4.8x10^-4)/(8x10^-6)
V₂=60V

q₁=q₂=4.8x10^-4C, V₁=240V and V₂=60V (Ans)