Question

A student connects a 21.0 V battery to a capacitor of unknown capacitance. The result is that 52.8 µC of charge is stored on the capacitor. How much energy (in J) is stored in the capacitor?

Answer 1

1.108 ×
`10^(-3)`J

Step-by-step explanation:

v=21.0v

Q=52.8×
`10^(-6)`

E=?

V=E/Q

E=v ×Q

=21 ×52.8 ×
`10^(-6)`

=1108.8 ×
`10^(-6)`

E= 1.108 ×
`10^(-3)`J