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.

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The electric field at a point 4.0 cm from the symmetry axis of the two surfaces is due to the charge on the cylindrical surface and the charge on the coaxial surface. The electric field due to the charge on the cylindrical surface is given by:$$E_1 = \frac{\lambda}{2\pi\epsilon_0 r}$$where:* $\lambda$ is the charge per unit length on the cylindrical surface* $\epsilon_0$ is the permittivity of free space* r is the distance from the point to the cylindrical surfaceThe electric field due to the charge on the coaxial surface is given by:$$E_2 = \frac{\lambda}{2\pi\epsilon_0 r}$$where:* $\lambda$ is the charge per unit length on the coaxial surface* $\epsilon_0$ is the permittivity of free space* r is the distance from the point to the coaxial surfaceThe total electric field at the point is the sum of the electric fields due to the charge on the cylindrical surface and the charge on the coaxial surface:$$E = E_1 + E_2 = \frac{\lambda}{2\pi\epsilon_0 r} + \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2\lambda}{2\pi\epsilon_0 r}$$Substituting the given values into the equation, we get:$$E = \frac{2(20 \times 10^{-9} \text{ C/m})}{2\pi(8.85 \times 10^{-12} \text{ F/m})(0.04 \text{ m})} = 0.45 \text{ kN/C}$$Therefore, the magnitude of the electric field at a point 4.0 cm from the symmetry axis of the two surfaces is 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