Question

Initially the potential difference of a 8 μF capacitor is 30 V. Then it is changed to 40 V. What is the increase in energy?

a) 28 × 10-4 J

b) 18 × 10-4 J

c) 8 × 10-4 J

d) 24 × 10-4 J

Answer 1

• C=8uF
• V1=30V
• V2=40V

Let Energies be E1 and E2

We know

$\boxed{\sf E=\dfrac{1}{2}CV^2}$

$\\ \sf\longmapsto E_1=\dfrac{1}{2}\times 8\times 30^2$

$\\ \sf\longmapsto E_1=4\times 900$

$\\ \sf\longmapsto E_1=3600J$

And

$\\ \sf\longmapsto E_2=\dfrac{1}{2}\times 8\times 40^2$

$\\ \sf\longmapsto E_2=4\times 1600$

$\\ \sf\longmapsto E_2=6400J$

Now

$\boxed{\sf E_{net}=E_2-E_1}$

$\\ \sf\longmapsto E_{net}=6400-3600$

$\\ \sf\longmapsto E_{net}=2800J$

$\\ \sf\longmapsto E_{net}=0.28\times 10^{-4}J$