How Electric Field Varies With Distance: Key Graphs
Physics · Electric Charges And Fields · NEET
Electric field falls off differently for each source: a point charge gives E proportional to 1/r squared, a dipole gives E proportional to 1/r cubed (faster fall), an infinite line gives E proportional to 1/r, and an infinite sheet gives a constant E (no fall). Memory hook: count the "spread" — point spreads in 3D so 1/r squared, dipole has two opposite charges that partly cancel so it drops one power faster (1/r cubed), and a big flat sheet cannot spread so E stays flat.
Field vs distance for each NEET source: dipole (1/r cubed) drops fastest, then point charge (1/r squared), then line (1/r); an infinite sheet stays constant, and a charged shell is zero inside R then falls as 1/r squared outside.
Your doubts, answered
Why does a dipole field fall as 1/r cubed but a point charge only as 1/r squared?
A single point charge has nothing to cancel it, so its field just spreads out and drops as 1/r squared. A dipole is a +q and a -q close together. From far away their fields almost cancel. The leftover field is what survives, and this leftover drops one extra power of r, giving 1/r cubed. So dipole field dies faster with distance. This is why NEET 2022 asked: field of two opposite charges at R much greater than L varies as 1/R cubed.
How does the field vary for a charged hollow sphere or conducting sphere?
Inside the sphere (r less than R) the field is exactly zero, because all the charge sits on the surface and the inside is empty of net field. At the surface it jumps up to a maximum. Outside (r greater than R) it behaves like a point charge at the centre, so E is proportional to 1/r squared and decreases as r grows. The graph is a flat zero line, then a jump, then a smooth 1/r squared fall.
Does distance matter for an infinite charged sheet?
No. For an ideal infinite plane sheet, E = sigma / (2 epsilon-zero), which has no r in it at all. The field is the same near the sheet and far from it, so the graph is a horizontal line. This is because the sheet is so large that moving away does not let the field spread out.
What is the field vs distance rule for an infinite line charge?
For a long straight line of charge, E = lambda / (2 pi epsilon-zero r), so E is proportional to 1/r. It falls slower than a point charge (which is 1/r squared). Reason: the line spreads its field in only 2D (like a cylinder), not full 3D, so it weakens more gently with distance.
Is the field inside a conducting sphere really zero, or just small?
It is exactly zero, not just small, in the static case. In electrostatics all the charge moves to the outer surface and the interior field cancels completely. So on a graph the field is a flat zero line all the way from the centre up to the radius R.
⚠️ The NEET trap ✗ Every electric field must decrease as you move away, so the field of an infinite sheet also gets weaker with distance. ✓ An ideal infinite charged sheet gives a constant field E = sigma / (2 epsilon-zero) that does not depend on distance. Only point charges (1/r squared), dipoles (1/r cubed) and lines (1/r) fall with r. 🧠 Match the source to its power of r BEFORE picking a graph: point = 1/r squared, dipole = 1/r cubed, line = 1/r, sheet = constant, inside a shell = zero.
Real NEET questions
2022
Two point charges -q and +q are placed at a distance L apart. The magnitude of the electric field intensity at a distance R (R much greater than L) varies as:
A · 1/R squared
B · 1/R cubed ✓
C · 1/R to the power 4
D · 1/R to the power 6
Solution: Two equal and opposite charges a small distance L apart form an electric DIPOLE. For a dipole, at large distance R (R much greater than L), the field is E = (1 / 4 pi epsilon-zero) x (p / R cubed) on the axis (and half of this magnitude on the equator, but both scale the same way). The key point: the two opposite fields partly cancel, killing one power of R compared to a single point charge. Point charge would be 1/R squared; the dipole is one power faster, so E is proportional to 1/R cubed. Answer: B.
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: By Gauss's law, take a spherical Gaussian surface. For r < R (inside the hollow sphere) the enclosed charge is zero, so E = 0. For r > R (outside), all charge Q looks like a point charge at the centre, so E = (1 / 4 pi epsilon-zero) x (Q / r squared), which is proportional to 1/r squared and therefore DECREASES as r increases. So: zero inside, then decreasing outside. Answer: B.
2023
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 squared ✓
B · V / r
C · rV / R squared
D · R squared V / r cubed
Solution: Surface potential of a charged sphere: V = kQ / R, so kQ = V R. Field outside (r > R) is like a point charge: E = kQ / r squared. Substitute kQ = V R to get E = V R / r squared = RV / r squared. This shows E is proportional to 1/r squared outside, matching the point-charge graph. 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.
Which source has the fastest falling electric field?
Among the standard NEET cases, the dipole falls fastest as distance grows, because E is proportional to 1/r cubed. Next is the point charge (1/r squared), then the infinite line (1/r), and the infinite sheet does not fall at all (constant).
What is the electric field inside a charged spherical shell?
It is exactly zero everywhere inside the shell, because the enclosed charge is zero (all charge sits on the surface). The field only exists outside, where it behaves as 1/r squared.
Why is the infinite sheet field independent of distance?
Because the sheet is treated as infinitely large, moving away does not give the field lines room to spread out. The lines stay parallel, so E = sigma / (2 epsilon-zero) is constant. In real problems this holds only for distances small compared to the sheet size.
How do I quickly pick the right graph in an exam?
First identify the source (point, dipole, line, sheet, or shell). Then recall its power of r: point 1/r squared, dipole 1/r cubed, line 1/r, sheet constant, shell zero inside then 1/r squared outside. Match that shape to the given curve.
