1. Feb 10, 2012

DracoMassakin

So the problem is "An electric field of 7.85*10^5 V/m is desired between two parallel plates, each of area 38.0 cm^2 and separated by 2.45 mm of air. What charge must be on each plate?"

Relevant equations: V=-E*d, C= (E0*A)/d, C=Q/V

I have calculated the voltage by taking V=-E*d=(-7.85*10^5 V/m)0.00245m= -1923.25V

Then I found the capacitance of the plates C= (E0*A)/d , (E0= Vacum Permativity) and got

C=((8.85*10^-12)0.0038 m^2)/0.00245m= 1.373*10^-11 F

But for the final calculation to find the charge I took Q= C*V = (1.373*10^-11 F)*-1923.25 V= -2.641*10^-8C So do I need to include the negative in that final calculation to get a negative charge or do I leave it out?

2. Feb 10, 2012

SammyS

Staff Emeritus
Hello DracoMassakin. Welcome to PF !

When we say there is a charge of Q on a capacitor what do we mean?

3. Feb 11, 2012

DracoMassakin

It means that it is the charge on each of the parallel plates right?

4. Feb 11, 2012

SammyS

Staff Emeritus
No.

If we have a capacitor with capacitance, C, in a circuit, and it is charged so the potential difference from one plate to the other is V, then we say the charge on the capacitor is Q, where Q = CV. In that case we actually have a charge of Q on one plate, and -Q on the other.