Question 122: Electrostats & Capacitance

Question

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V , its capacitance is: (in pF )

Accepted Answer

At first, a capacitor with capacitance $ C_0 = 100 $ pF is charged to a voltage $ V_0 = 60 $ V. This means it stores a charge $ Q = C_0 V_0 $.
Then, the battery is removed, so the total charge stays the same. Now, a second uncharged capacitor, $ C $, is connected in parallel. The two capacitors will now share the charge.
The final voltage across both capacitors is given in the question as 20 V. The charge is now shared between both capacitors:
Total charge = $ (C_0 + C) 	imes 20 $
But total charge stays the same as before, so:
$ (C_0 + C) 	imes 20 = C_0 	imes 60 $
Divide both sides by 20:
$ C_0 + C = 3 C_0 $
So, $ C = 2 C_0 $
Since $ C_0 = 100 $ pF, the second capacitor $ C $ is $ 2 	imes 100 = 200 $ pF.

Answer

600

Answer

100

Answer

400

Answer

200

Electrostats & Capacitance

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V , its capacitance is: (in pF )

Choose the correct answer: