Question

A 2uF capacitor is charged to a potential
difference of 200 V and then isolated. When it is connected across a second uncharged capacitor, the common potential difference becomes 40 V. What is the capacitance of the second capacitor?
A. 2uF
B. 4uF
С. 8uF
D. 8uF
E. 16uF​

Answer 1

Answer:

Initially the charge on the 2uF capacitor is Qi = CV = 2uF x 200V = 400uC =0.4mC

Finally after isolation when it is connected to uncharged capacitor, the amount of charge remains constant.

Using conservation of charge, we get

Qi = Q1 + Q2 = C1V1 + C2V2 = (2uF x 40V) + (C2 x 40V)

400uC = 80uC + (C2 x 40V)

320uC/40V = C2

8uF = C2

The capacitors capacitance is 8uF

Answer 2

Answer:

The capacitance of the second capacitor is equal to 8μF.

therefore, the option (C) is correct.

Explanation:

Given, the capacitance of 1st capacitor C₁ = 2μF

The potential difference applied across only on 1st capacitor V = 200V

Consider that the capacitance of 2nd capacitor is C₂.

The charge of 1st capacitor Q$_{i}$ = C₁V

Q$_{i}$ = 2μF× 200V

Q$_{i}$ = 400μC

When the 1st capacitor connected to 2nd capacitor, their common potential difference, V₁ = V₂= 40V

Because the 2nd capacitor is uncharged, the charge will be remain constant.

$Q_{i}=Q_{1}+Q_{2}$

C₁V = C₁V₁ + C₂V₂

400μC = (2μF)×(40V) + C₂×(40V)

400μC = 80μC + C₂×(40V)

C₂×(40V) = 320μC

$C_{2}=\frac{320\mu F}{40V}$

C₂ = 8μF

Therefore, the capacitance of the second capacitor will be 8μF.