Concept explainers
(a)
The electric potential of each of the three shells.
(a)
Answer to Problem 89P
The electric potential
Explanation of Solution
Given:
The inner shell is uncharged.
The middle shell has a positive charge
The outer shell has a negative charge
Formula used:
The expression for Gauss’s Law is given by,
The expression for potential is given by,
Calculation:
The required diagram is shown in Figure 1.
Figure 1
Applying Gauss’s law to a spherical Gaussian surface of radius
And
Applying Gauss’s law to a spherical Gaussian surface of radius
The potential between
And since
Therefore,
The inner shell too doesn’t carry charge, so the field between the points
Conclusion:
Therefore, The electric potential
(b)
The electric potential of each shell and the final charge on each shell.
(b)
Answer to Problem 89P
The electric potential of the shell
Explanation of Solution
Calculation:
Given:
The inner shell is uncharged.
The middle shell has a positive charge
The outer shell has a negative charge
Formula used:
The expression for Gauss’s Law is given by,
The expression for potential is given by,
Calculation:
When the inner and outer shells are connected their potentials become equal as a consequence of the distribution of charge.
The charges on the surface
The value of
The potential of shells a and c is given by,
The electric field between the points
The enclosed charge for
The potential difference between b and ciscalculated as,
Solve further,
Equate equation (2) and (3).
Substitute
The potential across
Conclusion:
Therefore, the electric potential of the shell
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Chapter 23 Solutions
Physics for Scientists and Engineers
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