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Applications of Gauss' Theorem > Electric Field due to a Uniformly Charged Thin Spherical Shell

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Estimated time: 9 minutes
CISCE: Class 12

Introduction

  • Consider a thin spherical shell of radius R, carrying a uniformly distributed charge Q (or surface charge density σ), such that Q = 4πR2σ.
  • Due to spherical symmetry, the electric field at any point depends only on the distance r from the centre, and is directed radially.
  • A Gaussian surface — an imaginary concentric sphere of radius r — is chosen to exploit this symmetry.

CISCE: Class 12

Point Outside the Shell

Case 1: Point Outside the Shell (r > R)

Step 1: Choose a Gaussian sphere of radius r > R, concentric with the shell.

Step 2: By symmetry, \[\vec{E}\] is uniform in magnitude over this surface and points radially outward.

Step 3: Apply Gauss's Law:

\[\oint\vec{E}\cdot d\vec{A}=E(4\pi r^2)=\frac{Q}{\varepsilon_0}\]

Step 4: Solve for E:

E = \[\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\]
Key Insight: Outside the shell, it behaves exactly like a point charge Q placed at the centre.
CISCE: Class 12

Point On the Surface

Case 2: Point On the Surface (r = R)

Substituting r = R:

E = ​\[\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}=\frac{\sigma}{\varepsilon_0}\]
This is the maximum field value, marking the boundary between zero (inside) and decaying (outside) field regions.
CISCE: Class 12

Point Inside the Shell

Case 3: Point Inside the Shell (r < R)

Step 1: Choose a Gaussian sphere of radius r < R, concentric with the shell.

Step 2: This Gaussian surface encloses no charge since all the charge resides on the shell's surface.

Step 3: Applying Gauss's Law:

E(4πr2) = \[\frac {0}{ε_0}\] ⟹ E = 0
Key Insight: The field is exactly zero everywhere inside a uniformly charged spherical shell.
CISCE: Class 12

Comparison Table — Field Regions

Region Condition Formula Field Behaviour
Inside shell r < R E = 0 No field; independent of r
On surface r = R E = \[\frac {σ}{ε_0}\] Maximum, finite value
Outside shell r > R E = \[\frac {1}{4πε_0}\frac {Q}{r^2}\]​ Decreases as 1/r2; acts as point charge
CISCE: Class 12

Graph: E vs r

Chart Suggestion (E–r graph):

  • X-axis: distance from centre r
  • Y-axis: electric field E
  • Behaviour: E = 0 for 0 ≤ r < R; sharp rise to peak value σ/ε0 at r = R; smooth 1/r2 decay curve for r > R.
CISCE: Class 12

Example

Given: A metal spherical shell of radius R = 0.25 m carries a charge Q = 0.2 μC.

Find: Electric field (a) inside the shell, (b) just outside the shell, (c) at r = 3.0 m.

Solution:

  1. Inside (r < R): E = 0
  2. Just outside (r = R):
    E = \[\frac {1}{4πε_0}\frac {Q}{R^2}\] = 2.88 × 104 N C−1
  3. At r = 3.0 m:
    E = \[\frac {1}{4πε_0}\frac {Q}{r^2}\] = 200 N C−1
CISCE: Class 12

Real-Life Analogy

  • Gravitational Shell Theorem Analogy: A uniformly charged spherical shell behaves like a hollow gravitational shell — just as gravity is zero inside a hollow planet-like shell, the electric field is zero inside a uniformly charged shell.
  • Everyday example: The interior of a charged metallic sphere (like a Van de Graaff generator dome) is field-free, which is why it is safe to stand inside a charged hollow conducting sphere.
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