Question

Derive an expression of electric potential at a point distance r from isolated unit postive charge

Answer 1

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

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 q and  q  already assembled at a point say A  

From the very definition of potential, you are bringing a charge say q0 and  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 x   as it is being bought from ∞

 Now,

Fconservative=kqq0x2

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

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°

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

dW=Fconsdx

dW=Fconsdx

dW=kqq0x2dx

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=∫∞rkqq0x2dx

W=kqq0∫r∞x−2dx

W=kqq0∫∞rx−2dx

W=kqq0[−1x]r∞

W=kqq0[−1x]∞r

W=−kqq0r

W=−kqq0r

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

VA=kqr

VA=kqr

Hence the potential at point A   is

VA=1/4πϵ0qr