Question 244: Electrostats & Capacitance

Question

A 10 $$\mu $$F capacitor is fully charged to a potential
difference of 50 V. After removing the source
voltage it is connected to an uncharged
capacitor in parallel. Now the potential
difference across them becomes 20 V. The
capacitance of the second capacitor is :

Accepted Answer

Initially,
Charge on capacitor 10 ÃÂ¼F
Q = CV = (10 ÃÂ¼F) (50V)
Q = 500 ÃÂ¼C
Final Charge on 10 ÃÂ¼F capacitor
Q = CV = (10 ÃÂ¼F) (20V)
Q = 200 ÃÂ¼C
From charge conservation,
Charge on unknown capacitor
Q = 500 ÃÂ¼C Ã¢ÂÂ 200 ÃÂ¼C = 300 ÃÂ¼C
$$ \Rightarrow $$ Capacitance (C) = $${Q \over V}$$ = $${{300} \over {20}}$$ = 15 $$\mu $$F

Answer

20 $$\mu $$F

Answer

15 $$\mu $$F

Answer

10 $$\mu $$F

Answer

30 $$\mu $$F

Electrostats & Capacitance

A 10 $\mu $F capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :

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