Ch 23: Electric Potential

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 23: Electric Potential

Problem 41d

Chapter 23, Problem 41d

A metal sphere with radius $r_a$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b$ . There is charge $+q$ on the inner sphere and charge $-q$ on the outer spherical shell. Use $E_{r} = [1 / (4 π ϵ_{0})] (q / r^{2})$ and the result from part (a) to find the electric field at a point outside the larger sphere at a distance $r$ from the center, where $r > r_{b}$ . Note: Part (a) asked to calculate the potential $V(r)$ for (i) $r < r_a$ ; (ii) $r_a < r < r_b$ ; (iii) $r > r_{b}$ . (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite..

Verified step by step guidance

Step 1: To calculate the potential V(r) for different regions, start by considering the potential due to a point charge. The potential at a distance r from a point charge q is given by V(r) = q / (4πϵ₀r).

Step 2: For region (i) r < r_a, the potential inside a conductor is constant and equal to the potential at its surface. Therefore, V(r) = V(r_a) = q / (4πϵ₀r_a).

Step 3: For region (ii) r_a < r < r_b, the potential is the sum of the potentials due to the inner sphere and the outer shell. The potential due to the inner sphere is V_inner = q / (4πϵ₀r), and the potential due to the outer shell is constant because the shell does not affect the potential inside it. Thus, V(r) = q / (4πϵ₀r) + constant.

Step 4: For region (iii) r > r_b, the potential is the sum of the potentials due to both spheres, but since the outer shell has charge -q, it cancels the effect of the inner sphere at distances greater than r_b. Therefore, V(r) = 0.

Step 5: To show the potential difference V_ab = q / (4πϵ₀) (1/r_a - 1/r_b), use the expression for the potential at the surface of each sphere and subtract the potential at r_b from the potential at r_a. This gives V_ab = V(r_a) - V(r_b) = q / (4πϵ₀) (1/r_a - 1/r_b).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Potential

Electric potential, V, is the work done per unit charge in bringing a positive test charge from infinity to a point in space. It is a scalar quantity and is measured in volts. In this problem, the potential at a point is the sum of potentials due to the inner and outer spheres, and it is zero at infinity.

Gauss's Law

Gauss's Law relates the electric field E to the charge distribution. It states that the electric flux through a closed surface is proportional to the enclosed charge. For spherical symmetry, it simplifies the calculation of electric fields, allowing us to determine E(r) using the symmetry of the charge distribution on the spheres.

Electric Field

The electric field, E, is a vector field representing the force per unit charge exerted on a test charge at any point in space. It is derived from the gradient of the electric potential, E = -∂V/∂r. In this problem, E is calculated for regions inside, between, and outside the spheres, considering the charge distribution.

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Textbook Question

A metal sphere with radius $r_{a}$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_{b}$ . There is charge $+ q$ on the inner sphere and charge $- q$ on the outer spherical shell. Calculate the potential $V (r)$ for (i) $r < r_{a}$ ; (ii) $r_{a} < r < r_{b}$ ; (iii) $r > r_{b}$ . (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite.

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Textbook Question

A metal sphere with radius $r_a$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b$ . There is charge $+q$ on the inner sphere and charge $-q$ on the outer spherical shell. Show that the potential of the inner sphere with respect to the outer is $V_{a} b = q / (4 π ϵ_{0}) (1 / r_{a} - 1 / r_{b})$ .

A very large plastic sheet carries a uniform charge density of $negative 6.00$ nC/m² on one face. As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential?

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Textbook Question

A metal sphere with radius $r_a$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b$ . There is charge $+q$ on the inner sphere and charge $-q$ on the outer spherical shell. Use $E_{r} = - \partial V / \partial r = (- \partial / \partial r) (1 / (4 π ϵ_{0}) q / r) = [1 / (4 π ϵ_{0})] (q / r^{2})$ and the result from part (a) to show that the electric field at any point between the spheres has magnitude $E (r) = [V_{a} b / (1 / r_{a} - 1 / r_{b})] (1 / r^{2})$ . Note: Part (a) asked to calculate the potential $V(r)$ for (i) $r < r_a$ ; (ii) $r_a < r < r_b$ ; (iii) $r > r_b$ . (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite..

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Textbook Question

Suppose the charge on the outer sphere is not $- q$ but a negative charge of different magnitude, say $- Q$ . Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different. Note: Part (a) asked to calculate the potential $V(r)$ for (i) $r < r_a$ ; (ii) $r_a < r < r_b$ ; (iii) $r > r_b$ . (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite. Part (b) asked to show that the potential of the inner sphere with respect to the outer is $V_ab=q/(4πϵ_0 ) (1/r_a -1/r_b)$ . Part (c) asked to use $E_r=-∂V/∂r=(-∂/∂r) (1/(4πϵ_0 ) q/r)=[1/(4πϵ_0 )](q/r^2)$ and the result from part (a) to show that the electric field at any point between the spheres has magnitude $E(r)=[V_ab/(1/r_a -1/r_b )](1/r^2)$ . Part (d) asked to use $E_r = [1/(4πϵ_0 )](q/r^2)$ and the result from part (a) to find the electric field at a point outside the larger sphere at a distance $r$ from the center, where $r > r_b$ .

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