A semicircle of radius a is in the first and second quadrants, with the center o

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 uniformly around the left half of the semicircle, and negative charge ?Q is distributed uniformly around the right half of the semicircle in the following figure.

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 the variables Q, a, constant ?, and electric constant ?0.

Part B

What is the direction of the net electric field at the origin produced by this distribution of charge?

a) +x direction

b) ?x direction

c) +y direction

d) ?y direction

e) another direction

Explanation / Answer

E_x = k*/a int_(pi/2)^(pi) cos(theta) dtheta - k*/a int_(0)^(pi/2) cos(theta) dtheta

E_y = k*/a int_(pi/2)^(pi) sin(theta) dtheta - k*/a int_(0)^(pi/2) sin(theta) dtheta = 0

So just worry about the x-component

E_x = k*/a int_(pi/2)^(pi) cos(theta) dtheta - k*/a int_(0)^(pi/2) cos(theta) dtheta

-2*k*/a; = charge/length; = Q/(pi*a/2)

-2*k*[Q/(pi*a/2)]/a = -4k*Q/(pi*(a^2))

k = 1/(4*pi*)

E_x = -4[1/(4*pi*)]*Q/(pi*(a^2))

E_x = Q/[(pi^2)*(a^2)*]

part b

the direction is along the +x-axis (Answer a) because the vertical components cancel out

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