A very long nonconducting solid cylinder of radius R = 0.0200 m has uniform volume charge density ρ = 4.00 × 10⁻⁶ C m⁻³. Neglect end effects. Derive E(r) for r R, then calculate E at r = 0.0100 m and r = 0.0400 m. Find V(0) − V(R), and sketch the qualitative E–r graph through r = 2R. Use ε₀ = 8.854 × 10⁻¹² F m⁻¹.

Question

A very long nonconducting solid cylinder of radius R = 0.0200 m has uniform volume charge density ρ = 4.00 × 10⁻⁶ C m⁻³. Neglect end effects. Derive E(r) for r < R and r > R, then calculate E at r = 0.0100 m and r = 0.0400 m. Find V(0) − V(R), and sketch the qualitative E–r graph through r = 2R. Use ε₀ = 8.854 × 10⁻¹² F m⁻¹.

Accepted Answer

Inside E = ρr/(2ε₀); outside E = ρR²/(2ε₀r). E(0.0100 m) = E(0.0400 m) = 2.26 × 10³ N C⁻¹. V(0) − V(R) = 45.2 V. Choose a coaxial Gaussian cylinder of length L. Cylindrical symmetry makes E radial and constant on its curved surface; flux through the flat ends is zero. For r < R, E(2πrL) = ρπr²L/ε₀, so E = ρr/(2ε₀). At 0.0100 m this is 2.26 × 10³ N C⁻¹. For r > R, the enclosed charge is ρπR²L. Therefore E = ρR²/(2ε₀r), giving 2.26 × 10³ N C⁻¹ at 0.0400 m. Potential difference is V(0)−V(R) = ∫₀ᴿE(r)dr = ρR²/(4ε₀) = 45.2 V. Potential falls outward along the outward field. The E–r graph starts at zero, rises linearly to ρR/(2ε₀) at R, stays continuous there, then falls as 1/r. A jump would incorrectly imply a surface-charge sheet added at R.

Field and potential of a charged solid cylinder

Choose your AP course · School year 2026–27 · May 2027 exam

What a complete response should do

Your self-assessment