A long cylindrical volume contains a uniformly distributed charge of density ρ Cm⁻³. The electric field inside the cylindrical volume at a distance x=(2 ε₀)/ρ m from its axis is ________ Vm⁻¹.

Solution

Physics sir · 4 steps

Draw a coaxial Gaussian cylinder of radius x: only its curved side carries flux, so E falls right out.

1. Choose the surface. Take a coaxial cylinder of radius x and length L inside the charge. By symmetry E points radially outward and has the same value everywhere on the curved surface. The flat end caps carry no flux because E runs parallel to them.
2. Apply Gauss's law. Flux through the curved surface is E × 2π x L. The charge enclosed is ρ × π x² L.

   E × 2π x L = (ρ π x² L)/ε₀ ⇒ E = (ρ x)/2ε₀

3. Put in x. With x = 2ε₀/ρ, you get E = ρ/2ε₀ × 2ε₀/ρ = 1.
4. So. The field at that point is 1 V m⁻¹.

The common slip: 2. Taking the curved area as π x L instead of 2π x L gives E = (ρ x)/ε₀, which is 2, not 1.

The idea to keep. Inside a uniformly charged long cylinder, E = (ρ r)/2ε₀, growing linearly with r. For a sphere it is (ρ r)/3ε₀. The geometry sets the number in the denominator.

Answer from the NTA final answer key. Solution by PrepWiser.