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1
Three point charges are located on a circular arc as shown in the figure below. (Take r = 4.04 cm. Let to the right be the +x direction and up along the screen be the +ydirection.)


a) What is the total electric field at P, the center of the arc?

E with arrow = _ i hat + _ j hat

(b) Find the electric force that would be exerted on a -4.77�nC point charge placed at P.

F with arrow = _ i hat + _j hat
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Added Fri, 28 Aug '15
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Solutions
1
The horizontal components of electric field due to charges q_1 ,q_2 and _q_3
E_x = E_1x + E_2x +E_3x

E_x = (kq_1 / r^2 )cos30^0 + (kq_2 /r^2) cos180^0 + (kq_3 / r^2 )cos30^0

Here r = radius of the circular arc = 4*10^-2 m
charges q_1 = q_3 = 3.0*10^-9 N and q_2 = -2.0*10^-9 C
k = 9*10^9 N.m^2 /C^2
plug all the values we get

E_x = 1.46*10^4 N/C - 1.125 *10^4 N/C + 1.46*10^4 N/C
E_x = 1.795*10^4 N/C

The verticle components of electric field at point p
E_2y = 0
As shown in the diagram E_1y and E_3y are cancells each other
E_y = 0
hence net electric field at the centre P E = sqrt ( (E_x)^2 + (E_y)^2) = 1.795*10^4 N/C
direction towards right
(b)
The force on the charge charge -5.00 nC
F =E *q = ( 1.795*10^4 N/C )(5.00*10^-9 C) = 8.975*10^-5 N
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Added Fri, 28 Aug '15
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