Question

Charge of uniform density (80 nC/m3) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical surfaces of radii 1.0 mm and 3.0 mm. Determine the magnitude of the electric field at a point which is 2.0 mm from the symmetry axis.

Answer 1

Answer: Magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.

Explanation:

Given: Density = 80 $nC/m^{3}$ (1 n = $10^{-9}$ m) = $80 \times 10^{-9} C/m^{2}$

$r_{1}$ = 1.0 mm (1 mm = 0.001 m) = 0.001 m

$r_{2}$ = 3.0 mm = 0.003 m

r = 2.0 mm = 0.002 m (from the symmetry axis)

The charge per unit length of the cylinder is calculated as follows.

$\lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})$

Substitute the values into above formula as follows.

$\lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})\\= 80 \times 10^{-9} \times 3.14 \times [(0.003)^{2} - (0.001)^{2}]\\= 2.01 \times 10^{-12} C/m$

Therefore, electric field at r = 0.002 m from the symmetry axis is calculated as follows.

$E = \frac{\lambda}{2 \pi r \epsilon_{o}}$

Substitute the values into above formula as follows.

$E = \frac{\lambda}{2 \pi r \epsilon_{o}}\\= \frac{2.01 \times 10^{-12} C/m}{2 \times 3.14 \times 0.002 m \times 8.85 \times 10^{-12}}\\= 18.08 N/C$

Thus, we can conclude that magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.