Question

State Gauss’s Theorem and use it to find the electric field due to an infinitely long charged wire or a thin spherical shell.

Hint:

Use Gauss’s law when symmetry exists: - Cylindrical symmetry → Use cylindrical Gaussian surface - Spherical symmetry → Use spherical Gaussian surface Also remember: electric field inside a charged spherical shell is always zero.

Answer 1

Concept: Gauss’s Theorem (Gauss’s Law) relates the electric flux through a closed surface to the charge enclosed by that surface. It is especially useful for systems with high symmetry such as spherical, cylindrical, or planar distributions. Mathematically, \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where:

• \( \vec{E} \) = electric field
• \( d\vec{A} \) = area vector
• \( Q_{\text{enc}} \) = charge enclosed
• \( \varepsilon_0 \) = permittivity of free space

Gauss’s law simplifies calculations when symmetry ensures that the electric field is constant over the Gaussian surface.
Part 1: Electric field due to an infinitely long charged wire Let the wire have linear charge density \( \lambda \) (charge per unit length).
Step 1: Choosing Gaussian surface Because of cylindrical symmetry, choose a cylindrical Gaussian surface of:

• Radius \( r \)
• Length \( L \)

The electric field is:

• Radially outward
• Same magnitude everywhere on curved surface

Step 2: Electric flux Flux passes only through the curved surface (not the flat ends): \[ \text{Flux} = E \times (2\pi rL) \]
Step 3: Charge enclosed \[ Q_{\text{enc}} = \lambda L \]
Step 4: Apply Gauss’s law \[ E(2\pi rL) = \frac{\lambda L}{\varepsilon_0} \] Cancel \( L \): \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]
Result: \[ \boxed{E = \frac{\lambda}{2\pi \varepsilon_0 r}} \] The electric field decreases inversely with distance from the wire.
Part 2: Electric field due to a thin spherical shell Let total charge on the shell be \( Q \) and radius be \( R \).
Case 1: Outside the shell (\( r>R \))
Step 1: Gaussian surface Choose a concentric sphere of radius \( r \).
Step 2: Flux \[ \text{Flux} = E \times 4\pi r^2 \]
Step 3: Enclosed charge \[ Q_{\text{enc}} = Q \] Applying Gauss’s law: \[ E(4\pi r^2) = \frac{Q}{\varepsilon_0} \] \[ \boxed{E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2}} \] Thus, the shell behaves like a point charge at the center.
Case 2: Inside the shell (\( r<R \))
Step 1: Enclosed charge No charge is enclosed: \[ Q_{\text{enc}} = 0 \] From Gauss’s law: \[ E = 0 \]
Result: \[ \boxed{E = 0 \quad \text{inside a charged spherical shell}} \]
Final Results:

• Infinitely long wire: \( E = \dfrac{\lambda}{2\pi \varepsilon_0 r} \)
• Outside spherical shell: \( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{Q}{r^2} \)
• Inside spherical shell: \( E = 0 \)