Question

What is the energy stored, in units of nanoJoules, on a 14.8 nF capacitor when the voltage applied across the capacitor is 7.2 V?

Answer 1

Answer:

The energy stored will be  3.83616e-7 J

Explanation:

Since the energy stored is given by

E = 1/2 CV²

So putting the values

E = 1/2 * 1.48e-8 * (7.2)²

E = 3.83616e-7 Joules

Answer 2

Answer:

383.6 nJ

Explanation:

The energy stored in a capacitor is given by the formula:

$E=\frac{1}{2}CV^2$

where

C is the capacitance

V is the voltage applied

In this problem, we have

C = 14.8 nF is the capacitance of the capacitor

V = 7.2 V is the voltage

Substituting into the equation, we find:

$E=\frac{1}{2}(14.8 nF)(7.2 V)^2=383.6 nJ$