18[E]) When a power source like a battery is connected to a circuit, an electric

Question

18[E]) When a power source like a battery is connected to a circuit, an electric field pushes electrons within the conducting wires causing them to flow. The electric field exists due to an uneven build up of charge on the surfaces of the wires (called a surface charge gradient). In this set of problems we will explore a simple model of how this all comes about and emphasize the connection between source charges, electric fields, and voltage differences in the context of electric circuits.

A very small segment of the conducting wire in a circuit can be modeled as a circular ring of radius 1 mm. Consider two such rings surrounding the z-axis separated by a small distance 2 mm. Let's first suppose that both rings have the same charge density. What is the strength of the electric field on the z-axis at the midpoint between the two rings?

Correct, computer gets: 0 V/m

19[E]) If you used a voltmeter to measure the voltage difference between the centers of the two rings, what measurement would you read?

Correct, computer gets: 0 V

20[H]) Now suppose that a power source is plugged into the circuit and that after a brief moment, you measure the voltage difference between the centers of the two rings to be 1 mV. What is the difference between the charge densities on the two rings?

21[E]) What's the strength of the average electric field between the centers of the two rings?

22[H]) What is the strength of the electric field at the midpoint between the centers of the two rings?

I just need solutions from 20 to 22 , dont worry about the workings . please make sure its right

Explanation / Answer

20 since the same current flows in the circuit , the charge density on the two rings will be same

hence difference between the charge densities = 0

21

Electric field = potential difference / d = 1 x 10-3 / 2 x 10-3 = 0.5 V/m

22. Electric field still cancel out at the midpoint , hence electric field at midpoint = 0

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