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.

 

[asteriatic]

Based on your calculations, it seems that the charge on each plate should be -2.641*10^-8 Coulombs. It is important to include the negative sign in your final calculation, as it indicates the direction of the charge. In this case, the negative sign indicates that the charge on each plate is negative, which is necessary to create the desired electric field between the plates.