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.(Figure 1)

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

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 +0 al

Explanation / Answer

Electric field due to positive charge is directed away from the charge and electric field due to negative charge is directed towards the charge.

Consider the quarter circle containing positive charge. ( In the second quadrant)

Charge per unit length = Q/(2*pi*a/4)=2*Q/(pi*a)

Consider a small segment of length dx at an angle t with y axis.

In terms of infinitisimal angle dt, dx=a*dt

Charge on this small segment =dq= charge per unit length * length of the segment= 2*Q*dt/pi

Distance of his segment from center =a

Then electric field magntidue = k*charge/ distance ^2

Where k is coloumb's constant

Electric field due to dq=dE=2*k*Q*dt/(pi*a^2)

Where t varies from 0 to pi/2.

Component of field along X axis =dE*sin(t)

=2*k*Q*sin(t)*dt/(pi*a^2)

Integrating from t =0 to t=pi/2, we get

Electric field along X axis =2*k*Q/(pi*a^2)

Component of field along Y axis =- dE*cos(t)

Electric field along Y axis = integration of -2*k*Q*cos(t)*dt/(pi*a^2)

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

Similar analysis for the portion in first quadrant will show thatelectric field along X axis due to negative charge =2*k*Q/(pi*a^2)

Electric field along Y axis = 2*k*Q/(pi*a^2)

So component along Y axis cancel each other .

Total field is along +ve X axis.

Total field magntidue =4*k*Q/(pi*a^2)

Part 2.

Option A is correct as total field is along +X direction.

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