When a power source like a battery is connected to a circuit, an electric field

Question

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

How would you figure out this problem?

Explanation / Answer

formula for electric field due to ring of radius R at disatcne z is

E = KQz/(z^2 + R^2)^3/2

where k is 9e8

z = 1mm

R = 0 mm

now direction of electric field due to one ring is towards positive z axis and direction due to other ring is towards negative z axis

so we have magnitude of electric field same for both the rings and net electric field become zero as electric field due to both rings cancel out each other

Now both rings have same charge density so charge on each ring will be same.

also we have same distance of point from both the rings.

E = 0 at mid point is the answer

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