Options:
A.
B.
C.
D.
E.
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Earlier discussion
1 older comment kept for context.
Collins
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 = C₁V
Q = 2μF× 200V
Q = 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.
C₁V = C₁V₁ + C₂V₂
400μC = (2μF)×(40V) + C₂×(40V)
400μC = 80μC + C₂×(40V)
C₂×(40V) = 320μC
C₂ = 8μF
Therefore, the capacitance of the second capacitor will be 8μF.