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The electrons in a conductor are. charge carriers would feel force and drift as long as the electric field is not zero. The charges distribute themselves so that the electric field is zero everywhere inside the conductor when there is no current inside or on the surface of the conductor. Inside a conductor, the electrostatic field is nil.
Now talking about the electric potential due to charged solid sphere, let us consider a charged sphere that has a symmetrical charge distribution. The electric field outside the sphere, according to Gauss’ Law, is the same as that produced by a point charge. This means that the potential outside the sphere is the same as the potential from a point charge. Consider a solid insulating sphere with a radius R and a charge distributed uniformly throughout its volume. Both the electric field and the electric potential outside the sphere are identical to the field and potential from a point charge.
When Gauss’ law is applied to the electric field of a charged sphere, the electric field environment beyond the sphere is found to be similar to that of a point charge. As a result, the potential is identical to that of a point charge:
Because the electric field inside a conducting sphere is zero, the potential at the surface remains constant:
The voltage inside a conductor at equilibrium is bound to be constant at the value it achieves at the conductor’s surface because the electric field is equal to the rate of change of potential. The charged conducting sphere serves as a nice illustration, but the idea is valid for all conductors in equilibrium.
In the sphere itself, what about it? We know the field within the sphere is zero if it is a conductor. What about its potential? The potential varies by an amount when one moves from a point on the outside to a location inside the sphere:
ΔV = -∫ E • ds
Given that E = 0, we can only infer that V is also zero, meaning that V is constant and equal to the potential at the sphere’s outer surface.
What takes place within the sphere? Due to the presence of a field inside the sphere, the potential is no longer constant. We demonstrated using Gauss’ Law that the field within an evenly charged insulator is as follows:
E =k Q r/R3
If all of the charge inside the sphere were concentrated at its centre, it would have the same potential as the vacuum at that place if it were conducting. To understand this, first note that the conducting sphere is a surface that must necessarily be equipotential. It follows that the field outside the sphere is radial and inverse-square just as if the charge had been at the centre due to the uniqueness of solutions to Laplace’s equation.
Applying Gauss’s formula now outside the sphere reveals that the electric field’s strength must be ?/4??0?2 because the total enclosed charge is merely the sphere’s internal charge. Integrating from infinity results in the same potential, ?/4??0?, because the electric field outside the sphere is the same as the one produced by a single charge Q at the centre without a conducting shell. Keep in mind that this reasoning is independent of the charge’s location within the sphere.
If the sphere is not conducting, the potential at the sphere will be determined by the conventional formula, ?/4??0? , where d is the separation between the charge and the specific point on the sphere. Keep in mind that the charge’s placement in this scenario is important; if the charge is not in the centre, the sphere will not be an equipotential surface. The volume of an infinitely thin shell is zero, hence its polarisation has no bearing on anything, regardless of whether the sphere is a dielectric or not.
We hope this short article on Electric Potential Due To Charged Solid Sphere has been helpful. Stay tuned to Embibe for such informative articles. Happy learning!
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