Question

What would be the potential at infinity due to an isolated charge?

Answer 1

Answer: Very simple Answer

Explanation: Electric potential can be defined in two ways,

It is the work done by an external agent at a point to bring a unit positive charge from infinity to that point.

It is the negative of the work done by the conservative force( which in this case is the electrostatic force) to bring a unit positive charge from infinity to that point.

VA=−W∞−>A,c.forceq0

We will use the second definition to derive an expression for an electric potential due to an isolated point charge.

Now suppose that you have a charge qalready assembled at a point say A. From the very definition of potential, you are bringing a charge say q0 from infinity to a point say B situated at a distance r from A. As the charge is bought from ∞ to B, the conservative force due to charge q opposes the work done by external agent to move q0 from ∞ to B. Let that force be labelled as Fconservative. Let the distance between the charge q and q0 beyond point B be xas it is being bought from ∞.

Now,

Fconservative=kqq0x2

Let there be a small displacement dx due to the external agent. In order to oppose the electrostatic force due to q, some work must be done.

Let a small work dWcons be done to displace charge q0 by dx.

Therefore,

dWcons=Fcons∙dx

Since x is decreasing, apart from taking θ from the dot product, we will take a negative sign for dx as well.

dW=Fcons(−dx)cos180° (as direction of displacement is oppsite to the conservative force)

dW=Fconsdx

dW=kqq0x2dx

We have to now find the work done by conservative force, for that we have to integrate both sides taking upper and lower limits as r and ∞ respectively.

W=∫r∞kqq0x2dx

W=kqq0∫r∞x−2dx

W=kqq0[−1x]r∞

W=−kqq0r

Potential is the negative of the work done by the conservative force per unit charge, therefore

VA=kqr

Hence the potential at point A is

VA=14πϵ0qr