Charge of uniform density (20 nC/m²) is distributed over a cylindrical surface (radius = 1.0 cm), and a second coaxial surface (radius = 3.0 cm) carries a uniform charge density of −12 nC/m². Determine the magnitude of the electric field at a point 4.0 cm from the symmetry axis of the two surfaces.

Question

Quizplus · Accepted Answer

The electric field at a point 4.0 cm from the symmetry axis can be found using Gauss's law. Since the point is outside both cylindrical surfaces, the electric field is determined by the net charge enclosed by a Gaussian surface that includes both cylinders. The charge on the inner cylinder is $Q_1 = \sigma_1 \cdot 2\pi r_1 L = 20 \times 10^{-9} \cdot 2\pi \cdot 0.01 \cdot L$, and the charge on the outer cylinder is $Q_2 = \sigma_2 \cdot 2\pi r_2 L = -12 \times 10^{-9} \cdot 2\pi \cdot 0.03 \cdot L$. The net charge is $Q_{net} = Q_1 + Q_2$, and the electric field at a distance $r = 0.04$ m from the axis is given by $E = \frac{Q_{net}}{2\pi \epsilon_0 r L}$. Substituting the values and solving gives an electric field magnitude of 0.45 kN/C.

Charge of Uniform Density (20 NC/m2) Is Distributed Over a Cylindrical

Charge of Uniform Density (20 NC/m²) Is Distributed Over a Cylindrical