Dielectric Capacitor Question

[sankara]

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Hi,

There's a passage in the TPR workbook where

Battary= 12V
Capacitor= 2uF
dielectric= 4

There's 3 parts to this passage
set-up: no dielectric, battery is connected to capacitor
Exp 1) dielectric, battery is connected to capacitor
Exp 2) dielectric, battery is disconnected to capacitor

one question asks " how much work is required to insert the dielectric in experiment 1.

I used PE= QV
since q increased by a factor of 4 I got (8 X 10uF)(12V)= 9.6 x 10^-5

The actual solution said Work=change in PE= 1/2 CV^2= 4.3 x 10 ^-4

I guess my problem is when dealing with a capacitor question I do not know when to use PE= qV v. 1/2 CV^2

or in other words, why could I have not used PE=QV

Cheers!

 

[sankara]

anyone?

 

[joshto]

for capacitors, PE=1/2qv