A parallel plate capacitor with a capacitance of 5µF is connected to a 500 V batter. Calculate the charge on the plates and the energy stored.

Question

asked Mar 11, 2020 81.3k views

1 Answer

Ben Strombeck · Answer 1 · 2020-03-17T14:50:55+0000

Step-by-step explanation:

It is given that,

Capacitance,
$C=5\ \mu F=5* 10^(-6)\ F$

Voltage, V = 500 volts

Charge on the plates,
Q=C* V

$Q=5* 10^(-6)\ F* 500$

Q = 0.0025 C

Energy stored in the capacitor is given by :

E=(1)/(2)CV^2

$E=(1)/(2)* 5* 10^(-6)\ F* (500\ V)^2$

E = 0.625 Joules

Hence, this is the required solution.