If the electric field is zero, is the potential zero too?
Short answer: E = -dV/dr. Zero field means V is constant, not zero. Inside, V equals the surface value kQ/R.
The trap
The wrong idea
“Where the electric field is zero, the potential must also be zero.”
It sounds reasonable, and a lot of JEE/NEET aspirants carry it into the exam. Here's a one-question check, then the explanation that makes it click.
Quick check
Try it before you peek
Inside a charged hollow conducting sphere, the potential is:
- Zero
- Constant, equal to surface value
- Increasing inward
Show the answer
B. Constant, equal to surface value
E = -dV/dr. Zero field means V is constant, not zero. Inside, V equals the surface value kQ/R.
The click
Field and potential are linked through a rate of change, not by value: E = −dV/dr (in general, E = −∇V). The field tells you how fast the potential changes with position. So E = 0 in a region means V is constant there, but that constant can be any value.
Inside a charged hollow conducting sphere of radius R carrying charge Q, the field is zero. By Gauss's law, a closed surface drawn inside encloses no charge, and by symmetry the field there must vanish. Since E = 0, V cannot change as you move around inside, so V everywhere inside equals its value at the surface: V = kQ/R, where k = 1/(4πε₀). That is not zero unless Q = 0.
A graph of V against r makes it clear: V is flat at kQ/R from the centre out to r = R, then falls as kQ/r outside. The field is the negative slope of this graph: zero on the flat part and kQ/r² outside.
The reverse trap exists too: V = 0 doesn't mean E = 0. At the midpoint between charges +q and −q, V = 0 (with V taken as zero at infinity), yet the field there is strong and points from + toward −. At the midpoint between two equal positive charges, E = 0 but V is positive. Remember also that where V = 0 is a choice of reference; only differences in V are physical.
Got this one? The Knot Scan finds the misconception you don't know you have, in 8 questions.