Electric Field Inside vs Outside a Charged Conducting Sphere
Physics · Electric Charges And Fields · NEET
For a charged conducting sphere of radius R with charge Q, the field is zero everywhere inside (r < R), then it drops off as E = kQ/r^2 outside (r > R). At the surface the field jumps suddenly from 0 to a maximum value E = kQ/R^2. Memory hook: "Zero inside, cliff at the surface, then a 1/r^2 slide down."
E-r graph for a charged conducting sphere: flat zero inside (r < R), a sudden jump to the maximum kQ/R^2 at the surface, then a 1/r^2 decline outside.
Your doubts, answered
Is the electric field really zero everywhere inside a charged conducting sphere?
Yes. All the extra charge on a conductor sits on the outer surface, so any Gaussian surface drawn inside (r < R) encloses zero charge. By Gauss's law, E = 0 for every point with r < R, not just the centre. This is true whether the sphere is solid or hollow, because the charge only lives on the outside skin.
Why is the field the largest exactly at the surface?
Just inside, E = 0. Just outside, the full charge Q acts as if it were at the centre, so E = kQ/R^2. Since R is the smallest value of r for which the outside formula applies, kQ/R^2 is the biggest value E ever reaches. As r grows beyond R, E only gets smaller. So the surface is the peak of the graph.
Does a hollow sphere behave differently from a solid conducting sphere?
No. For a conductor, charge always moves to the outer surface, so a solid conducting sphere and a hollow conducting shell of the same radius and charge give the exact same field: zero inside and kQ/r^2 outside. The inside being empty (hollow) or filled (solid metal) does not change the result.
How is a conducting sphere different from a uniformly charged insulating (solid) sphere?
In an insulator the charge is spread through the whole volume, so inside (r < R) the field is not zero. It grows linearly, E = kQr/R^3, from 0 at the centre up to kQ/R^2 at the surface. Outside, both cases are identical: E = kQ/r^2. NEET loves to mix these two graphs up, so read whether the sphere is 'conducting' or 'uniformly charged solid'.
What happens to the field right at r = R? Is it one value or two?
The field is discontinuous at the surface. It jumps from 0 (inside) to kQ/R^2 (outside). This sudden step is why the E-r graph has a vertical line at r = R. The commonly quoted 'surface value' kQ/R^2 is the value just outside the surface.
⚠️ The NEET trap ✗ The field is zero at the centre and slowly builds up to a maximum as you move to the surface from inside. ✓ For a CONDUCTING sphere the field is zero at EVERY interior point (r < R), not just the centre. It only becomes non-zero the instant you cross the surface, jumping straight to kQ/R^2. 🧠 'Builds up inside' is the insulator graph (E = kQr/R^3). For a conductor, inside is flat zero. Always check the word: conductor = flat zero, solid insulator = rising line.
Real NEET questions
NEET 2019
A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre
A · increases as r increases for r < R and for r > R
B · is zero for r < R, decreases as r increases for r > R ✓
C · is zero for r < R, increases as r increases for r > R
D · decreases as r increases for r < R and for r > R
Solution: Step 1: The sphere is a metal (conductor), so all charge sits on the outer surface. Step 2: For r < R, a Gaussian sphere encloses zero charge, so by Gauss's law E = 0 inside. Step 3: For r > R, the full charge Q acts as if at the centre, so E = kQ/r^2. This means E is proportional to 1/r^2, which decreases as r increases. Therefore E = 0 inside and decreases outside. Answer: B.
NEET 2023 Phase 2
If a conducting sphere of radius R is charged, then the electric field at a distance r (r > R) from the centre of the sphere would be (V = potential on the surface of the sphere):
A · RV / r^2 ✓
B · V / r
C · rV / R^2
D · R^2 V / r^3
Solution: Step 1: Surface potential of a charged conducting sphere is V = kQ/R, so kQ = VR. Step 2: For r > R the sphere behaves as a point charge, so E = kQ/r^2. Step 3: Substitute kQ = VR: E = VR/r^2 = RV/r^2. Answer: A.
Solved Electric Charges And Fields NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the electric field inside a charged conducting sphere?
It is zero at every point inside (r < R), because all the charge sits on the surface and any interior Gaussian surface encloses no charge.
What is the field just outside the surface of the sphere?
E = kQ/R^2 (with k = 1/4πε0). This is the maximum value the field reaches, right at r = R.
How does the field vary outside the sphere?
It follows E = kQ/r^2, so it decreases as 1/r^2 as you move farther away, exactly like a point charge placed at the centre.
Is there a jump in the graph at the surface?
Yes. The E-r graph is flat at zero inside, then jumps suddenly to kQ/R^2 at r = R, then curves down as 1/r^2. The sudden step at the surface is a key exam point.
Does this apply to a hollow sphere too?
Yes. A hollow conducting shell and a solid conducting sphere of the same radius and charge give identical fields, since charge always resides on the outer surface.
