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A -2.34-nC point charge is located at the center of a conducting spherical shell. The shell has an inner radius of 2.00 m, an outer radius of 4.00 m, and a charge of 8.53-nC. (a) What is the magnitude of the electric field at r = 1.00 m?  N/C (b) What is the magnitude of the electric field at r = 3.00 m?  N/C (c) What is the magnitude of the electric field at r = 5.00 m?  N/C (d) What is the surface charge density, σ, on the outside surface of the shell? (You may enter your calculation using scientific notation.)  C/m2C/m2

Question

A -2.34-nC point charge is located at the center of a conducting spherical shell. The shell has an inner radius of 2.00 m, an outer radius of 4.00 m, and a charge of 8.53-nC. (a) What is the magnitude of the electric field at r = 1.00 m?  N/C (b) What is the magnitude of the electric field at r = 3.00 m?  N/C (c) What is the magnitude of the electric field at r = 5.00 m?  N/C (d) What is the surface charge density, σ, on the outside surface of the shell? (You may enter your calculation using scientific notation.)  C/m2C/m2

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Solution

(a) The electric field (E) inside a conductor in electrostatic equilibrium is zero. Therefore, at r = 1.00 m, which is inside the conducting shell, the electric field is 0 N/C.

(b) At r = 3.00 m, we are inside the conducting shell but outside the point charge. The electric field in this region is due to the point charge only. The formula for the electric field due to a point charge is E = kQ/r^2, where k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge. Substituting the given values, we get E = (8.99 x 10^9 N m^2/C^2)(-2.34 x 10^-9 C)/(3.00 m)^2 = -2.22 N/C. The negative sign indicates that the field is directed toward the charge.

(c) At r = 5.00 m, we are outside both the point charge and the conducting shell. The electric field in this region is due to both charges. The total charge is the sum of the charges of the point charge and the shell, which is -2.34 nC + 8.53 nC = 6.19 nC. Substituting this into the formula for the electric field, we get E = (8.99 x 10^9 N m^2/C^2)(6.19 x 10^-9 C)/(5.00 m)^2 = 2.23 N/C. The positive sign indicates that the field is directed away from the total charge.

(d) The surface charge density (σ) on the outside surface of the shell is the total charge on the shell divided by the surface area of the shell. The surface area of a sphere is 4πr^2. Therefore, σ = Q/(4πr^2) = (8.53 x 10^-9 C)/(4π(4.00 m)^2) = 1.68 x 10^-10 C/m^2.

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