Question

A capacitor of capacitance 5uF is charged to a potential of 100 V. Another capacitor of capacitance 10uF is charged to a potential of 150 V. The two capacitors are then joined with plates of like charges connected together.
(a) Find the total energy stored in the system before and after the capacitors are joined.
(b) What happens to the difference in energy in the two cases?

Answer 1

Given that,

Capacitor of capacitance $C_{1}= 5\ \mu F$

Potential $V_{1}= 100\ V$

Capacitor of capacitance $C_{2}=150\ \mu F$

Potential $V_{2}=150\ V$

Before capacitors joined,

We need to calculate the total energy

Using formula of energy

$E=E_{1}+E_{2}$

$E=\dfrac{1}{2}C_{2}V_{2}^2+\dfrac{1}{2}C_{2}V_{2}^2$

Put the value into the formula

$E=\dfrac{1}{2}\times5\times10^{-6}\times(100)^2+\dfrac{1}{2}\times10\times10^{-6}\times(150)^2$

$E=0.1375\ J$

We need to calculate the charge on first capacitor

Using formula of charge

$Q_{1}=C_{1}V_{1}$

Where, $C_{1}$ = first capacitor

$V_{1}$ = potential

Put the value into the formula

$Q_{1}=5\times10^{-6}\times100$

$Q_{1}=500\ \mu C$

We need to calculate the charge on second capacitor

Using formula of charge

$Q_{2}=C_{2}V_{2}$

Where, $C_{2}$ = second capacitor

$V_{2}$ = potential

Put the value into the formula

$Q_{2}=10\times10^{-6}\times150$

$Q_{2}=1500\ \mu C$

We need to calculate the common potential

Using formula of  potential

$V=\dfrac{total\ charge}{total\ capacitance}$

Put the value into the formula

$V=\dfrac{(500+1500)\times10^{-6}}{(5+10)\times10^{-6}}$

$V=133.3}\ V$

After capacitors joined,

We need to calculate the total energy

Using formula of energy

$E'=\dfrac{1}{2}(C_{1}+C_{2})V^2$

Where, V = common potential

Put the value into the formula

$E'=\dfrac{1}{2}(5+10)\times10^{-6}\times(133.3)^2$

$E'=0.1332\ J$

(b). We need to calculate the difference in energy

Using formula of difference in energy

$E''=E-E'$

Where, E = final energy before capacitors joined

E' = final energy after capacitors joined

Put the value into the formula

$E''=0.1375-0.1332$

$E''=0.0043\ J$

Hence, (a). The final energy before capacitors joined is 0.1375 J

The final energy after capacitors joined is 0.1332 J.

(b). The difference in energy in the two cases is 0.0043 J.