Question

Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

Answer 1

$C_1=20.0pF,$

$C_2=50.0pF$

(a) Charge on both capacitor is equal

So, $q=\left(\frac{C_1\times C_2}{C_1+C_2}\right)\times V$

$=\left(\frac{20\times 50}{20+50}\right)\times 6.0pC$

$V_1-\frac{q}{C_1}=\frac{20\times 50}{70\times 20}\times 6.0$

$=1.71V$

$V_2=(6.00-1.71)V=4.29V$

(b) $E_1=\frac{q_2}{2C_1}$

$={[\frac{20\times 50}{70}\times 6.0]}^2\times \frac{1}{2\times 20}$

$=184pJ$

$E_2=\frac{q_2}{2C_1}$

$={[\frac{20\times 50}{70}\times 6.0]}^2\times \frac{1}{2\times 50}$

$=73.5pj$