Physics · Electrostatic Potential And Capacitance · NEET
Field zero means the potential does not CHANGE, not that it is zero. E = -dV/dr, so E = 0 means V has no slope — it is flat. A flat line can sit at any height. Inside the shell that height equals the surface value kQ/R. So the potential is constant and non-zero, while the field is genuinely zero. Zero slope is not the same as zero value.
It is the same as everywhere inside and on the surface: V = kQ/R = Q/(4 pi epsilon0 R). The centre is not special. Because no work is done moving a test charge anywhere inside (field is zero), every interior point including the centre sits at the surface potential. Students often expect the centre to have a peak — it does not; the graph is flat.
No. For a HOLLOW conducting shell (or any surface charge), all charge is on the surface, field inside is zero, so V is CONSTANT = kQ/R inside. For a uniformly charged SOLID insulating sphere, there is charge throughout, field inside is not zero, and V rises toward the centre reaching V_centre = 3kQ/(2R). For NEET, read the word 'shell' or 'conducting' carefully — that means constant potential inside.
They are equal. Potential is CONTINUOUS across the shell surface — both give kQ/R at r = R. The field is what is discontinuous (it jumps from 0 inside to kQ/R^2 = sigma/epsilon0 outside). So on the V-r graph the flat inside line meets the falling outside curve smoothly with no gap at r = R.
A uniformly charged shell has spherical symmetry, so by Gauss's law the field outside is exactly kQ/r^2, identical to a point charge Q sitting at the centre. Integrating that field from infinity gives V = kQ/r for all r greater than or equal to R. This is why for any external point you can ignore the shell's size and treat Q as concentrated at the centre.
A thin spherical shell is charged by some source. The potential difference between two points C and P, both shown inside the shell, is (Take 1/(4 pi epsilon0) = 9 x 10^9 SI units):
If a conducting sphere of radius R is charged, then the electric field at a distance r (r greater than R) from the centre would be (V = potential on the surface of the sphere):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Outside (r greater than or equal to R): V = kQ/r = Q/(4 pi epsilon0 r). Inside (r less than R): V = kQ/R = constant, equal to the surface value.
No. It is constant and equal to kQ/R, which is non-zero for a charged shell. Only the field is zero inside.
The potential is maximum throughout the interior and at the surface (all equal to kQ/R). It then decreases as 1/r outside. There is no single peak point; the whole inside is at the maximum.
Potential is continuous across the surface (both sides give kQ/R). The field is discontinuous — it jumps from 0 inside to kQ/R^2 outside.
A conducting or hollow shell has constant potential inside (kQ/R). A uniformly charged solid insulator has potential rising to 3kQ/(2R) at its centre because the field inside is not zero.
