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Ch 23: Electric Potential
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 23, Problem 41b

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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Step 1: To solve part (a), we need to calculate the potential V(r) in three regions: (i) inside the inner sphere (r < r_a), (ii) between the inner sphere and the outer shell (r_a < r < r_b), and (iii) outside the outer shell (r > r_b). Use the formula for the potential due to a point charge, V(r) = q/(4πϵ_0 r), and consider the contributions from both the inner sphere and the outer shell.
Step 2: For part (b), calculate the potential difference between the inner sphere and the outer shell, V_ab. Use the formula V_ab = q/(4πϵ_0) (1/r_a - 1/r_b), which represents the potential difference due to the charge distribution on the spheres.
Step 3: In 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 has magnitude E(r) = [V_ab/(1/r_a - 1/r_b)](1/r^2). This involves differentiating the potential function with respect to r.
Step 4: For part (d), find the electric field at a point outside the larger sphere (r > r_b) using E_r = [1/(4πϵ_0)](q/r^2). This requires applying the formula for the electric field due to a point charge and considering the charge distribution on the spheres.
Step 5: In part (e), consider the case where the charge on the outer sphere is -Q instead of -q. Show that the potential difference V_ab and the electric field between the spheres remain unchanged, but the electric field outside the larger sphere is affected by the change in charge magnitude. This involves re-evaluating the potential and electric field expressions with the new charge configuration.

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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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Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is essential for calculating the electric field in symmetric charge distributions, such as spherical shells. Here, it helps determine the electric field between and outside the spheres based on the charge distribution.
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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, the field is calculated between and outside the spheres using the potential difference and 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. 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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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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Textbook Question

Certain sharks can detect an electric field as weak as 1.01.0 μμV/m. To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.51.5­V AA battery across these plates, how far apart would the plates have to be?

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