Electrical Capacitator and Electrical Field question. Please help

[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?

 

[SammyS]

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?

Hello DracoMassakin. Welcome to PF !

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

 

[DracoMassakin]

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

 

[SammyS]

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

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.