Potential Energy in an Electric Field: Learn How to Calculate For Charges and Dipole

Some approaches are necessary to simplify the calculations and derivations of the potential energy in an external field, and we list these below, aligning with your NCERT textbook. 

For this discussion, we need to know that the charge (s) and dipole are not creating this external field. Instead, your NCERT book asks to assume that “what is specified is the electric field E or the electrostatic potential V due to the external sources” (p. 58, Electrostatic Potential and Capacitance).
That gives us three conditions. 

The charge we bring does not disturb the existing field.

What this means that the test charge (q) is quite small. It cannot by any means alter how the external sources are distributed.
It’s more like a feather floating in the wind. Too light. And which does not change anything in the wind. 

Even when the charge isn’t tiny, the external sources are fixed.

When external sources are massive or are held together by other forces, it won’t matter if the charge is finite (in other words, charge with magnitude and not infinitesimally tiny like a test charge).

It’s like a magnet placed near Earth. The magnet, with its field, is finite. But the Earth’s gravitation is too big to compare, and the magnet really doesn’t or cannot really affect Earth’s pull. 

Sources too far away can have a constant influence in interest region.

Sometimes the source charges can be literally too far away but can still create a uniform electric field. Placing a local charge in this field won’t affect anything. 

The goal here is to calculate the potential energy of the charge or a system of charges in such a field.
What we DON’T need is to find the energy that’s stored in the sources that created the electric field.
We NEED to study how the charge RESPONDS when placed in an external field. Not about how it originated.

To find the potential energy of a single charge placed in an external field, the logic, pretty much, follows the same old concept that you learnt in potential due to a point charge. And the only difference is that we are considering the field to be external here. 

To bring a charge from infinity to a point in an external field, there needs to be an effort. That effort is done by you as an external agent. That effort is against the electric field that’s already there or predefined. 

That’s basically how work is stored as potential energy. The equivalent in mathematical terms would be like this below.
We have a test charge as q and we place this one in an external field E. This field already has the electrostatic potential that we represent as V(r) at every point. 

If we’re to bring the charge from infinity, and we assume the field’s effect is zero at infinity and so the potential V is equal to zero. This charge is brought from infinity to a point r in the field.
This is done slowly and with no acceleration. The force we apply to move the charge from infinity must be able to balance the electrostatic force.
And we’ve established earlier in this chapter that the work done by an external agent in an electric field is, 

W = qV(r)

Then, the work gets stored as the potential energy at that point, which is,

U = qV(r)

Beyond the Scope of CBSE

There is one more aspect of the physics of force and potential energy we need to be sure of, which is the conservative nature. And the same will apply here and help you determine the sign convention more appropriately, which is extremely important when practising the NCERT Chapter 2 Physics Class 12 Solutions. 

So when a particle moves in a force field - whether gravitational or electric - there are two things happening at once

• The force does work on the particle
• The potential energy of the charge and the system goes through a change

Going by the work-energy theorem, the change in kinetic energy of a particle is exactly equal to the total work done on it by all the forces. 

When we’re considering a conservative force field like gravitational force or Coulomb’s force, the total mechanical energy is the sum of kinetic energy and potential energy, ie., E = K + U = constant. To maintain its constancy, a change in kinetic energy must or has to accommodate the change in potential energy. Or vice-versa. 

Now, if we’re taking a test charge that’s moving in an electric field, the electrostatic potential energy (U) definition that there is a movement so slow in order to maintain ΔK = 0. That means there isn’t any kinetic energy change, or it remains constant. So the work done by the external agent that causes the charge to move is stored as the potential energy in the system.
And we know the change in potential energy of a system when a particle moves from a to b positions within the conservative force field is the negative of the work done by the force during that displacement. That is, ΔU = - W_conservative.

When the force is conservative and the work done is towards the direction of movement or displacement of the particle, the kinetic energy increases. What do you think will happen with the potential energy to maintain the conservation of total mechanical energy (E=K+U)?
Now, if you are the external agent exerting a force against the electric field, much like taking an object against gravity to a higher point, the potential energy will be higher. 

So here is a way to memorise:

Move against the field → Store energy → Potential energy (U) increases.

Move with the field → The field is doing the work → Potential energy (U) decreases.  

 

For a single test charge brought from infinity to the external field somewhere at an arbitrary point r, we know the potential energy is the charge multiplied by the electric potential at that point, U = qV(r). 

But what happens when there are two charges brought into the field?

There will be two interactions (get ready for more calculations here!).

• Two of these charges will be interacting with the electric field at their own level. 
• Both charges will interact with each other too.

The approach here is to find the total potential energy of the system, and we do that by bringing one charge first and then the next. It is quite similar to the steps used to find the potential energy of a system of charges. 

The external field’s potential could be V_ext(r), while the two charges (q_1 and q_2) are brought from infinity to places in the external field of positions r_1 and r_2.  

Now, what we have are these steps:

1. Bringing the first charge q_1 to the external field to r_1 as the position

The first charge interacts with the field and we can use the above equation for a single charge. 

U_1 = q_1V_ext(r_1)

Don’t forget that this charge q_1 is creating its own electric field. 

2. Bringing the second charge q_2 to the external field to r_2 as the position

This second charge, q_2, is going to face two fields:

• One that’s created by q_1
• The other that’s already the external field that the first charge was put on 

By logic, we would define the second potential energy equation in the external field as

U_2_ext = q_2V_ext(r_2)

The field of q_1 will create a potential energy that will be acquired by q_2 at position r_2. And we will use the older equation for electrostatic potential due to a point charge 

V_q_1(r_2)  = k x q_1/ |r_2 - r_1|

k is what you can remember as the Coulomb constant with the equation 1/4πε_0

Now to find the potential energy of the interaction between the two charges q_1 and q_2, we get

U_12 = q_2 x V_q_1(r_2) = k x q_1q_2/|r_2 - r_1|

 

Total potential energy of the system

U_total = U_1 + U_2_ext + U_12

U_total = q1 x V_ext(r1) + q2 x V_ext(r2) + (k x q1 x q2 / |r2 - r1|)

 

To summarise, we can say that when two charges are brought into an external field, the total potential energy is the sum of:

• Energy of first charge q_1
• Energy of second charge q_2
• Energy of interaction between two charges q_1 and q_2

If there are three charges, it will be the sum of every separate interaction in pairs along the energy of each charge. 

 

So the dipole, as you dabbled in Class 12 electrostatics, is described as two opposing charges (+q and -q) separated by a distance (2a).

When placing this dipole in a uniform electric field, we get the dipole moment as p = q x 2a and it’s directed from -q to +q. What happens to the field here?

Obviously, there isn’t any force, but there would be torque. 

τ = p x E

The magnitude of this torque of the dipole is τ = pEsinθ

Now think of what would be the work done on the dipole, if you were to move it from the first angle θ_0 to θ_1.

This is possible when you apply an external torque to balance the field’s torque, so that there’s no angular acceleration?

The work done here would be

W = pE(cosθ_0 - cosθ_1)

We are using cosθ because we are calculating the work done on the dipole between two different orientations. 

So now we have to calculate the potential energy change between two orientations here and we know that the this stored energy is the negative of the work done if the angle between the two is 90 degrees which makes potential energy zero.

That is, we get,

U = - pEcosθ