Question

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

Answer 1

Answer:

A) |E| = $\frac{Q}{\pi^2e_{o}a^2 }$

B ) +X direction

Explanation:

A) Determine the magnitude of the net electric field at the origin produced by the distribution of charge

The net electric field is made up of the electric fields from both quadrants

for the first quadrant the electric field can be expressed as

E1 = $\frac{Q}{2\pi ^2e_{o}a^{2} } ( i+j)$  --- ( 1 )

we have to apply the principle that field lines point away from a positive charge to a negative charge to determine the electric field in the second quadrant

E2 = $\frac{Q}{2\pi ^2e_{o}a^{2} } ( i-j)$ ---- ( 2 )

when we add up equation 1 and equation 2

The magnitude of the net electric field =

|E| = $\frac{Q}{\pi^2e_{o}a^2 }$

B ) The direction of the net electric field is : + x-direction