Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 25, Problem 23P

Consider two conducting spheres with radii R1 and R2 separated by a distance much greater than cither radius. A total charge Q is shared between the spheres. We wish to show that when the electric potential energy of the system has a minimum value, the potential difference between the spheres is zero. The total charge Q is equal to q1 + q2, where q1 represents the charge on the first sphere and q2 the charge on the second. Because the spheres are very far apart, you can assume the charge of each is uniformly distributed over its surface. (a) Show that the energy associated with a single conducting sphere of radius R and charge q surrounded by a vacuum is UE = keq2/2R. (b) Find the total energy’ of the system of two spheres in terms of the total charge Q, and the radii and R1 and R2. (c) To minimize the energy, differentiate the result to part (b) with respect to q1 and set the derivative equal to zero. Solve for q1 in terms of Q and the radii. (d) From the result to part (c), find the charge q2. (e) Find the potential of each sphere. (f) What is the potential difference between the spheres?

(a)

Expert Solution
Check Mark
To determine

To show: The energy associated with a single conducting sphere is UE=keq22R .

Answer to Problem 23P

The energy associated with a single conducting sphere is UE=keq22R .

Explanation of Solution

Given info: The radii of two conducting sphere is R1 and R2 . The total charge shared between them is Q . The charge on first and second sphere is q1 and q2 respectively.

Write the expression to calculate the capacitance of a sphere of radius R .

C=Rke

Here,

ke is the Coulomb’s law constant.

Write the expression to calculate the potential difference.

ΔV=keqR

Here,

ΔV is the potential difference of the capacitor.

q is the charge on a single sphere.

Write the expression to calculate the energy stored in the capacitor.

UE=12C(ΔV)2

Substitute Rke for C and keqR for ΔV in above equation.

UE=12(Rke)(keqR)2=keq22R

Conclusion:

Therefore, the energy associated with a single conducting sphere is UE=keq22R .

(b)

Expert Solution
Check Mark
To determine
The total energy of the system of two spheres in term of q1 , Q and the radii R1 and R2 .

Answer to Problem 23P

The total energy of the system of two spheres in term of q1 , Q and the radii R1 and R2 is 12keq12R1+ke2(Qq1)2R2 .

Explanation of Solution

Given info: The radii of two conducting sphere is R1 and R2 . The total charge shared between them is Q . The charge on first and second sphere is q1 and q2 respectively.

Write the expression to calculate the capacitance of a sphere of radius R .

C=Rke

Write the expression to calculate the total energy of the system of two sphere.

UE=12q12C1+12q22C2

Substitute R1ke for C1 and R2ke for C2 in above equation.

UE=12q12(R1ke)+12q22(R2ke) (1)

The sum of charge of both sphere are,

Q=q1+q2q2=Qq1

Substitute Qq1 for q2 in above equation.

UE=12keq12R1+ke2(Qq1)2R2

Thus, the total energy of the system of two spheres in term of q1 , Q and the radii R1 and R2 is 12keq12R1+ke2(Qq1)2R2 .

Conclusion:

Therefore, the total energy of the system of two spheres in term of q1 , Q and the radii R1 and R2 is 12keq12R1+ke2(Qq1)2R2 .

(c)

Expert Solution
Check Mark
To determine
The value of q1 by differentiating the result of part (b).

Answer to Problem 23P

The value of q1 is q1=R1QR1+R2 .

Explanation of Solution

Given info: The radii of two conducting sphere is R1 and R2 . The total charge shared between them is Q . The charge on first and second sphere is q1 and q2 respectively.

The total energy of the system of two spheres in term of q1 , Q and the radii R1 and R2 is,

UE=12keq12R1+ke2(Qq1)2R2 .

Differentiate the above equation with respect to q1 and equate to zero.

dUEdq1=0ddq1[12keq12R1+ke2(Qq1)2R2]=0keq1R1+ke(Qq1)R2(1)=0q1=R1QR1+R2

Conclusion:

Therefore, the value of q1 is q1=R1QR1+R2 .

(d)

Expert Solution
Check Mark
To determine
The value of q2 from part (c).

Answer to Problem 23P

The value of q2 is q2=R2QR1+R2 .

Explanation of Solution

Given info: The radii of two conducting sphere is R1 and R2 . The total charge shared between them is Q . The charge on first and second sphere is q1 and q2 respectively.

The value of q1 is,

q1=R1QR1+R2 .

The sum of charge of both sphere are,

Q=q1+q2q2=Qq1

Substitute R1QR1+R2 for q2 in above equation.

q2=QR1QR1+R2=R2QR1+R2

Conclusion:

Therefore, the value of q2 is q2=R2QR1+R2 .

(e)

Expert Solution
Check Mark
To determine
The potential of each sphere.

Answer to Problem 23P

The potential of each sphere is keQR1+R2 .

Explanation of Solution

Given info: The radii of two conducting sphere is R1 and R2 . The total charge shared between them is Q . The charge on first and second sphere is q1 and q2 respectively.

Write the expression to calculate the potential of first sphere.

V1=keq1R1

Substitute R1QR1+R2 for q1 in above equation.

V1=keR1×R1QR1+R2=keQR1+R2

Write the expression to calculate the potential of second sphere.

V2=keq2R2

Substitute R2QR1+R2 for q1 in above equation.

V2=keR2×R2QR1+R2=keQR1+R2

Thus, the potential of each sphere is keQR1+R2 .

Conclusion:

Therefore, the potential of each sphere is keQR1+R2 .

(f)

Expert Solution
Check Mark
To determine
The potential difference between the spheres.

Answer to Problem 23P

The potential difference between the spheres is zero.

Explanation of Solution

Given info: The radii of two conducting sphere is R1 and R2 . The total charge shared between them is Q . The charge on first and second sphere is q1 and q2 respectively.

The potential difference is,

ΔV=V1V2

Substitute keQR1+R2 V1andV2 in above equation.

ΔV=keQR1+R2keQR1+R2=0

Conclusion:

Therefore, the potential difference between the spheres is zero.

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Chapter 25 Solutions

Physics for Scientists and Engineers

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