Problem: A semicircle of radius a is in the first and second quadrants, with the center of curvature at the origin.  Positive charge +Q is distributed around the left half of the semicircle (0 to 90 degrees) and negative charge -Q is distributed uniformly around the right half of the semicircle (90 to 180 degrees).Part A:  What is the magnitude of the net electric field at the origin produced by this distribution of charge?Express your answer in terms of variables Q, a, and appropriate constants.Part B:  what is the direction of the net electric field at the origin produced by the distribution of this charge?+X direction, -X direction, +Y direction, or -Y direction

FREE Expert Solution

There are two halves of a semicircle in this problem.

The electric field for the first half of the semicircle in the first quadrant with the negative charge is:

${\mathbit{E}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{2}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{a}}^{\mathbf{2}}}\mathbf{\left(}\stackrel{\mathbf{^}}{\mathbf{i}}\mathbf{+}\stackrel{\mathbf{^}}{\mathbf{j}}\mathbf{\right)}$

The electric field lines always point towards a negative charge, away from a positive charge.

This implies that the electric field will point away from the center of the charged half of the semicircle in the second quadrant.

The electric field in the second quadrant will have a positive x-component and a negative y component.

Problem Details

A semicircle of radius a is in the first and second quadrants, with the center of curvature at the origin.  Positive charge +Q is distributed around the left half of the semicircle (0 to 90 degrees) and negative charge -Q is distributed uniformly around the right half of the semicircle (90 to 180 degrees).

Part A:  What is the magnitude of the net electric field at the origin produced by this distribution of charge?

Express your answer in terms of variables Q, a, and appropriate constants.

Part B:  what is the direction of the net electric field at the origin produced by the distribution of this charge?

+X direction, -X direction, +Y direction, or -Y direction

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Electric Field Lines concept. You can view video lessons to learn Electric Field Lines. Or if you need more Electric Field Lines practice, you can also practice Electric Field Lines practice problems.