# What is the meaning of potential of the capacitor?

I am little confuse about the potential of the capacitor. Suppose we have two parallel plates of capacitor and both have equal and opposite charges on it. Let say plate "A" has positive charge and plate 'B' has negative charge. Now my teacher told me that they have potential V1 and V2 respectively. What does this potential of the plates mean?

In the above diagram the potential of the plate 'A' at any point is V1. Does that mean that V1 is the potential due to both plate 'A' and 'B' and similarly V2 is the potential due to both plate A and Plate B. Please someone explain this?

• Where is the potential considered to be zero? Jun 23, 2022 at 15:42
• At infinity i guess Jun 23, 2022 at 15:44

Yes. $$V_{1}$$ and $$V_{2}$$ represent the potential due to the entire charge distribution.

This is exactly the same as standard potential at a point in space.

The difference in this potential $$[V_{2}-V_{1}]$$ represents the amount of work required to move unit charge from plate $$V_{1}$$ to $$V_{2}$$

below is wrong, as conductive plates in the steady state are in equipotential as stated by John doty, was forgetting they are conductive and not fixed constant surface charge density!

I think your confusion may stem from "potential on the plate" instead of "potential at a point in space".

I would like to point out that the potential at a point in space along the plate is not constant.

However the formula E= v/d assumes the plates are infinite,and as such, the potential at any point on the plate is constant[do the math], and as such, we can then talk about the "potential of the plate"

• Since in the usual abstraction, we assume the plates are perfectly conductive, the potential in space at the surface of a plate is constant. But for a finite plate, the surface charge density isn't constant. Jun 23, 2022 at 15:14
• Yes you are correct! Jun 23, 2022 at 15:18
• You mean that if the plate is not consider to be infinite large than the potential on the different points along the plate would not be same, ie it will not be a equipotential surface anymore? Jun 23, 2022 at 15:47
• "However the formula E= v/d assumes the plates are infinite". The formula assumes the separation $d$ is much less than the dimensiond of the plate so that the field can be considered uniform. Jun 23, 2022 at 16:01
• @MdFaiyaz In electrostatic state, all conductors have equipotential surface. Since your capacitor is in electrostatic state, no matter what the size of plates is, plates will be equipotential. Jun 23, 2022 at 16:09

What does this potential of the plates mean?

Potential of a particular plate means the amount of work done in bringing a unit positive charge from infinity ($$V=0$$) to that particular plate. In your question, $$V_1$$ represents sole potential of plate A and $$V_2$$ represents sole potential of plate B. These potentials are independent of each other, if we ignore how capacitor is created i.e. using a battery or else.

In case of parallel plate capacitors the distance between the plates (d) is much smaller as compared to the length and breadth (area) of the plate. So when charged, these plates produces uniform electric field between the plates (ignoring the end effect).

Using relation, $$dV=-\vec{E}.\vec{dr}$$ $$V_{\text{negative}} - V_{\text{positive}}=-E\Delta r$$ (Path of integral is from positive plate to negative plate)

Therefore, $$\Delta V=E\Delta r$$

In case of parallel plate capacitor, electric field between the plates is: $$E = \frac{Q}{A \varepsilon_0}$$
Where $$Q$$ is charge on each plate, $$A$$ is area of plate. (Both plates have same area and charge i.e. A and Q respectively)

And
$$\Delta r = d = \text{Distance between the plates}$$
So finally, potential difference between the plates from relation $$\Delta V=E\Delta r$$ is: $$\Delta V = \frac{Qd}{A \varepsilon_0}$$

This potential difference ($$\Delta V$$) is often referred as potential difference across the capacitor and sometimes it is represented simply by $$V$$.