Potential Energy of a System of Charge: Class 12 Physics Notes

Electrostatic potential energy (U) is the work done by an external force needed to bring a charge from a reference point with zero potential energy (usually from infinity) to a certain point in an electric field. This work is always against the electric field's force. But it must be ensured the charge is moved slowly without bringing any change to its kinetic energy, and the applied external force has to balance the electrostatic force at every point.

The right method is to know how to find the total external work done. This work is for bringing charges from infinity to a point that we choose, isn't it!

Now, think that these charges have a separation that's infinite, and the result will help us fetch their final positions.

How many charges? For now, we can use names for three of them as q_1, q_2, and q_3 and use the final positions for each charge, respectively, as r_1, r_2, and r_3.

1. Bringing one charge q_1 from infinity to a specified position r_1

To bring the charge q_1 and to a position or point r_1, there is zero work done on the first charge. 

Now this position is in a system. You can remember these reasons why the work is initially zero.

In this step of bringing the first charge, we know there isn’t any other charge present to create a field. 

So when there is no external electric field, there cannot be any external energy that requires the exertion of any force.

The external force (F_ext) that’s there to move the charge is just enough to counter the electric force(F_E), which gives us F_ext = - F_E. And we know F_E = 0. So W_1 (work done on the first charge) equals zero. 

Now think of bringing other charges, one after another. For the second, there will be a non-zero work done. It's because it will move against the field and potential the first charge creates. 

The first charge creating the potential you previously learnt to derive with our potential due to a point charge Class 12 guide. 

V = k x q_1/r = 1/4πε_0 x q_1/r

2. Bringing the second charge q_2 to second position r_2 in the field of q_1

The next step is to bring the second charge, q_2, from infinity near q_1 to a position that has a distance of r_12 from q_1. 

The first charge, q_1, creates an electric potential, V_1. To find the work done on q_2, we must know the potential where it finally is, ie., at the final position of q_2.

Here work must be done against the electric field of q_1. 

The work equation will be the product of the charge and the potential difference between two points, like this way W = qΔV

This work relationship with potential and charge comes from the equation that details the electrostatic potential energy (U) and electrostatic potential (V), when we are looking at two points 

ΔV = ΔU/q

Also remember, the work done by an external force (which is equal and opposite to the electric force, F_ext = -F_E, to move a charge without acceleration is fully stored as the potential energy difference (ΔU)). That means, W_external = ΔU = U_final - U_initial

And since electrostatic potential, by definition, is generalised as the potential energy per unit charge, so we can say 

V = U/q 

Or we can say when moving between two points 

ΔV = W_ext/q 

W_ext = qΔV

Now you can think about what the equation would be for work done to bring the second charge q_2 to the field of the first charge with respect to distance, r_2.

This tells us the potential V_1 at the position r_2 depends on the distance from the source charge, q_1. This distance we define as r_12. 

V_1 = 1/4πε_0 x q_1/r_12

Substitute this for the work done by the second charge. 

W_2 = q_2V_1(r_2) = q_2 x q_1x1/4πε_0 x 1/r_12 

As the Coulomb force is conservative, the work done or the electrostatic potential energy of this system of two charges is dependent on the initial and final positions of the charge q_2. 

3: Bringing a third charge q_3 which is against the field of the two charges

We have the two charges q_1 and q_2 in a system and we know how to calculate the potential of the impact of the two. The potential energy is cumulative. That also means, if we have to bring another charge to this same system, the work done will be relatively more. 

V_1_2 is how we denote the potential at r_3 that arises because of q_1 and q_2. 

To obtain the third charge, q_3, and to understand the total impact, we must simplify and apply superposition (the same used when finding potential due to a system of charges).

The work done, W_3 is the third charge (q_3). This individual charge gets multiplied by the potential V_1_2 at the third position (r_3). And that's with respect to the two other charges and their positions would look like this

W_3 = q_3V_1_2(r_3) = 1/4πε_0 (q_1q_3/r_13 + q_2q_3/r_23)

Now, what would be the total potential energy?

It’s the total work done of the three system charges 

W_total = W_1 + W_2 +W_3…

ie.,

U_total = 1/4πε_0 (q_1q_2/r12 + q_1q_3/r_13 + q_2q_3/r_23)

No matter how you place them in any order, the total potential energy of the system of three charges would be the same. This tells us that the force is conservative.