A metal sphere with radius ra r_a is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rbr_b. There is charge +q+q on the inner sphere and charge −q-q on the outer spherical shell. Calculate the potential V(r)V(r) for (i) rr\u003erbr \u003e r_b. (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take VV to be zero when rr is infinite.

Ch 23: Electric Potential

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 23: Electric Potential

Problem 41a

Chapter 23, Problem 41a

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.

Verified step by step guidance

Step 1: Understand the problem setup. We have a metal sphere with radius r_a and charge +q, and a hollow metal spherical shell with radius r_b and charge -q. We need to calculate the potential V(r) in different regions and analyze the electric field.

Step 2: For part (a), calculate the potential V(r) for different regions. (i) For r < r_a, the potential inside a conductor is constant, so V(r) = V(r_a). (ii) For r_a < r < r_b, use the formula for the potential due to a point charge: V(r) = (1/(4πϵ_0)) * (q/r). (iii) For r > r_b, the potential is due to both charges, but since the net charge is zero, V(r) = 0.

Step 3: For part (b), calculate the potential difference V_ab between the inner sphere and the outer shell. Use the formula V_ab = q/(4πϵ_0) * (1/r_a - 1/r_b). This represents the potential difference due to the charges on the spheres.

Step 4: For part (c), use the expression for the electric field E_r = -∂V/∂r to show the magnitude of the electric field between the spheres. Substitute the potential difference V_ab and simplify to find E(r) = [V_ab/(1/r_a - 1/r_b)] * (1/r^2).

Step 5: For part (d), calculate the electric field at a point outside the larger sphere using E_r = [1/(4πϵ_0)] * (q/r^2). For part (e), consider the case where the charge on the outer sphere is -Q instead of -q. The potential difference V_ab remains the same, but the electric field outside the larger sphere changes due to the different net charge.

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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 is calculated for different regions around the spheres, considering the contributions from both the inner and outer spheres.

Gauss's Law

Gauss's Law relates the electric field to the charge distribution. It states that the electric flux through a closed surface is proportional to the enclosed charge. This principle helps in determining the electric field between and outside the spheres, as the symmetry of the spherical charge distribution simplifies calculations using this law.

Superposition Principle

The superposition principle states that the net potential at a point is the algebraic sum of potentials due to individual charges. This concept is crucial for calculating the potential V(r) in different regions, as it involves summing the potentials from the inner sphere and the outer shell, each contributing to the total potential at a given point.

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Related Practice

Textbook Question

How much excess charge must be placed on a copper sphere $25.0$ cm in diameter so that the potential of its center, relative to infinity, is $3.75$ kV? What is the potential of the sphere's surface relative to infinity?

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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 $E_{r} = [1 / (4 π ϵ_{0})] (q / r^{2})$ 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..

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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 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

Certain sharks can detect an electric field as weak as $1.0$ $μ$ V/m. To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary $1.5$ V AA battery across these plates, how far apart would the plates have to be?

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Textbook Question

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. The surface charge density for each plate has magnitude $47.0$ nC/m^2. If the separation between the plates is doubled while the surface charge density is kept constant at the given value, what happens to the magnitude of the electric field and to the potential difference?

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