Energy stored in parallel plate capacitor

[xamy]

Can anyone please tell me whether I have solved this question correctly or not.

1. Homework Statement
A parallel plate capacitor has a charge of 0.040µC on each plate with a potential difference of 240 V. The parallel plates are separated by 0.20 mm of air. What energy is stored in this capacitor?

Homework Equations

Q/V=eo A/d
Cair=Aeo/d
E=1/2 CV2

The Attempt at a Solution

Q/V=eo A/d

A=Qd/Veo

A=(0.040u)(0.20/1000)/(240)(8.85x10-12)

A=3.766x10-3 m2

Cair=Aeo/d

=[3.766x10-3(8.85x10-12)]/(0.20/1000)

=0.1666uF

E=1/2 CV2

=1/2(0.1666u)(240)2

=4.7994uJ

 

[gneill]

You have reached the correct result. But you took a pretty long route to get there. Note that the relationship between charge and potential on a capacitor is Q = C*V. You've been given both Q and V so you could have obtained C directly.

 

[xamy]

Alright thankyou!