Question

3.
A 6uF capacitor is charged from 10
volts to 20 volts. Increase in energy
will be
(A) 18 x 10-4 Joules
(B) 9x10-4 Joules
C) 4.5x10-4 Joules
(D) 9x10-6 Joules
​

Answer 1

Answer:

answer has to be $9*10^{-4}$ Joules

Explanation:

As they are asked for is the increase in energy you should apply $energy=1/2(CV^{2} )$ for both cases and then get the different,

Answer 2

Answer:

The difference in the energy is ΔE = 0.9 mJ.

Explanation:

Given,

Capacitance of a capacitor (C) = 6μC

$V_i = 10 volts\\V_f = 20 volts$

To find,

Increase in energy (ΔE)

We know that energy of a capacitor is $E = \frac{1}{2} CV^2$

So, $\Delta E = \frac{1}{2} C(V_f^2-V_i^2)$....(1)

We substitute the given data in the question in equation (1)

$\Delta E = \frac{1}{2}\times 6 \times 10^{-6} \times (20^2-10^2)$

$\implies \Delta E = 9\times10^{-4}J$

Therefore, the difference in the energy is ΔE = 0.9 mJ.