derive an expression for electric potential due to an isolated point charge
Answers
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
Explanation:
thus the the expression for electric potential due to an isolated point charge is explained....
