Physics · Electrostatic Potential And Capacitance · NEET
It is the same EVERYWHERE — at every point inside the solid and at every point on the surface. Reason: in the static state the field inside a conductor is zero (E = 0). Since E = -dV/dr, if E is zero then V does not change from point to point. So the whole conductor is one single equipotential. Students often think only the surface is equipotential; the inside is at that same value too.
The potential V is the SAME everywhere on the conductor. For a small, sharply curved region, think of it locally as a small sphere of radius r. Since V = kQ_local/r is fixed, a smaller r means the region holds charge in a way that gives a larger sigma. Roughly sigma is proportional to 1/r (1/radius of curvature). So sharp tips and small spheres carry high sigma and high field just outside — this is why lightning rods are pointed and charge 'leaks' from tips (corona discharge).
E = sigma / epsilon-0, directed normal (perpendicular) to the surface at that point. Note it is sigma/epsilon-0, NOT sigma/(2 epsilon-0). The 2 epsilon-0 result is for an isolated infinite sheet with field on both sides. For a conductor, the field exists only on the outside (inside is zero), so all the flux goes out on one side, giving the full sigma/epsilon-0.
No. V is a single number for the whole conductor — it does not vary from a high-sigma tip to a low-sigma flat face. sigma varies over the surface, but V is constant. Do not confuse the local quantity sigma (changes point to point) with the global quantity V (one value for the whole body).
Potential is set by the whole charge distribution and the geometry, not by the local sigma alone. The conductor's free electrons rearrange themselves precisely so that V comes out equal everywhere; that self-adjustment is what forces E = 0 inside. The uneven sigma is the PRICE the conductor pays to keep V uniform.
A conducting sphere of radius R is charged. The electric field at a distance r (r > R) from the centre of the sphere is (V = potential on the surface of the sphere):
Which of the following statements are correct? A. Inside a conductor, the electrostatic field is zero. B. Electric field at the surface of a charged conductor does not depend on its surface charge density. C. The interior of a charged conductor can have no excess charge in the static situation. D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point. E. The electrostatic potential is zero everywhere inside a charged conductor.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Because the field inside is zero, no work is needed to move a charge from any point to any other point inside or on the surface, so all points are at the same potential. The whole conductor is a single equipotential volume.
E = sigma / epsilon-0, pointing normal (perpendicular) to the surface, where sigma is the local surface charge density and epsilon-0 is the permittivity of free space.
At the most sharply curved parts — the pointed tips and small-radius regions. Roughly sigma is proportional to 1/(radius of curvature), so tips have high sigma, high field, and can cause corona discharge.
Yes, in the static situation E = 0 everywhere inside the conducting material. The free electrons rearrange until any internal field is cancelled.
For a sphere of radius R at surface potential V, the potential outside falls as V(r) = VR/r for r > R (like a point charge), while inside and on the surface it stays constant at V.
