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Ch 23: Electric Potential
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 23, Problem 41c

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. Use Er=V/r=(/r)(1/(4πϵ0)q/r)=[1/(4πϵ0)](q/r2)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)=[Vab/(1/ra1/rb)](1/r2)E(r)=[V_ab/(1/r_a -1/r_b )](1/r^2). Note: Part (a) asked to calculate the potential V(r)V(r) for (i) r<rar < r_a; (ii) ra<r<rbr_a < r < r_b; (iii) r>rbr > 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..

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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, so V(r) = (1/(4πϵ_0)) * (q/r - q/r_b).
Step 3: For part (b), calculate the potential difference V_ab between the inner sphere and the outer shell. Use the formula V_ab = V(r_a) - V(r_b) = q/(4πϵ_0) * (1/r_a - 1/r_b).
Step 4: For part (c), use the expression for the electric field E_r = -∂V/∂r = (1/(4πϵ_0)) * (q/r^2) and the result from part (a) to show that the electric field between the spheres is E(r) = [V_ab/(1/r_a - 1/r_b)] * (1/r^2).
Step 5: For part (d), use the expression for the electric field outside the larger sphere, 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. For part (e), if the charge on the outer sphere is -Q instead of -q, the potential difference V_ab and the electric field between the spheres remain the same, but the electric field outside the larger sphere changes due to the different charge distribution.

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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.
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Electric Potential

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.
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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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Related Practice
Textbook Question

How much excess charge must be placed on a copper sphere 25.025.0 cm in diameter so that the potential of its center, rela­tive to infinity, is 3.753.75 kV? What is the potential of the sphere's surface relative to infinity?

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Textbook Question

A metal sphere with radius 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) r<rar < r_a; (ii) ra<r<rbr_a < r < r_b; (iii) r>rbr > 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.

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Textbook Question

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. Use Er=[1/(4πϵ0)](q/r2)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 rr from the center, where r>rbr > r_b. Note: Part (a) asked to calculate the potential V(r)V(r) for (i) r<rar < r_a; (ii) ra<r<rbr_a < r < r_b; (iii) r>rbr > 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..

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Textbook Question

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. Show that the potential of the inner sphere with respect to the outer is Vab=q/(4πϵ0)(1/ra1/rb)V_ab=q/(4πϵ_0 ) (1/r_a -1/r_b).

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Textbook Question

A very large plastic sheet carries a uniform charge density of 6.00-6.00 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

Suppose the charge on the outer sphere is not q-q but a negative charge of different magnitude, say Q-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)V(r) for (i) r<rar < r_a; (ii) ra<r<rbr_a < r < r_b; (iii) r>rbr > 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. Part (b) asked to show that the potential of the inner sphere with respect to the outer is Vab=q/(4πϵ0)(1/ra1/rb)V_ab=q/(4πϵ_0 ) (1/r_a -1/r_b). Part (c) asked to use Er=V/r=(/r)(1/(4πϵ0)q/r)=[1/(4πϵ0)](q/r2)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)=[Vab/(1/ra1/rb)](1/r2)E(r)=[V_ab/(1/r_a -1/r_b )](1/r^2). Part (d) asked to use Er=[1/(4πϵ0)](q/r2)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 rr from the center, where r>rbr > r_b.

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