Question

Numerical: A capacitor of 200 picofarad is charged to a potential difference of 100 volts. It's plates are then connected parallel to another capacitor them potential difference in this combination fall to 60 volts. What is the capacitance of second capacitor?
(Working of this solution)

Answer 1

Answer:

The capacitance of the other capacitor is 1333.3 pF.

Explanation:

C' = 200 pF, V' = 100 V, C'' , V''= 0

Common potential, V = 60 V

Use the formula of the common  potential

$V = \frac{C' V' + C'' + V''}{C'+ C''}\\\\60 = \frac{200\times 100 + 0}{200 + C''}\\\\1200+ 6 C''= 2000\\\\C'' = 1333.3 pF$