# Potential difference between two assymetrically charged plates

What is the potential diff between two plates of charges $q_1$ and $q_2$ ($q_1$ $>$ $q_2$)? I know the equation is $V = \frac{q_1-q_2}{2C}$ where $C$ is the capacitance of the system but how has this equation been derived?

I apologize if this is a duplicate question. Google searches with different search terms didn't yield any answers.

Let us say plate A has a charge q1 and plate B, which faces plate A has a charge q2. By making use of the fact that the net field in the bulk of a conductor in static conditions is zero, and that the net field near the outer surface of a conductor equals [local surface charge density/€0], you can prove the following:

1. Charge on the outer surfaces of A and B are each (q1+q2)/2.
2. By conservation of charge, the charges on the facing surfaces of A and B are (q1-q2)/2 and -(q1-q2)/2 respectively.
3. Now using the result for net field near the outer surface of a conductor mentioned above, the field in the region between the plates is, E=([q1-q2)/2]/(A*€0), pointing towards plate B if q1>q2. Here A is the area of each plate.

Finally, potential difference is E*d, where d is the distance between the plates. This is valid because for infinitely large plates, the field in the region between the plates can be considered uniform. Since capacitance C=A€0/d, we obtain the potential difference V= (q1-q2)/2C

If the charges on the plates of the capacitor are +Q and -Q then the PD between the plates is V = Q/C where C is capacitance. (This is the definition of capacitance.)

If we add the same amount of charge Q'>Q to each plate, this has no effect on the PD between the plates, because it will increase the absolute potential of each plate by the same amount. The charges on the plates are then :
$q1 = Q'+Q$
$q2 = Q'-Q$.

From these we get :
$q1-q2 = 2Q$
$Q = \frac12(q1-q2)$.
Insert this into the formula for PD :
$V = \frac{Q}{C} = \frac{q1-q2}{2C}$.