Question

3)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?

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

Answer 1

The electric field along the axis of a charged ring is given by

,

where k=8.987×10⁹ N·m²/C² is the proportionality constant ,

R is the radius of the ring,

is the linear charge density

and z is the distance along the ring's axis at which the field is calculated.

At the midpoint between both rings (z=0), one ring is located at z=+1mm and the other one is located at z=-1 mm. the electric field is

The electric field is defined by .

Case 1) If both rings have the same charge "q", the directions of the fields are opposed, and since they have the same magnitude, the net electric field in the midpoint beween the rings is zero. Since the statement of the problem does not provide the vaulue of the charge density, this must be the case, because we can't determine the numeric magnitude of the field for the case 2.

Case 2) if one ring has a charge "q", and the other one has a charge "-q", one field woulds push and the other would pull a positive charge in the midpoint between them. In this case, the fields add up, and the resultant field is twice the field produced by a single ring at 1 mm along its axis.

for case 2, the electric field is

4) The voltage difference can be determined by

The distance is 2 mm, but since the magnitude of the field is 0 N/C, the voltage difference is 0 in this case.