# Electric field

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A charge of +10.7 micro-Coulombs is placed at the origin of a coordinate system. Another charge of -12.6 micro-Coulombs is placed at x = +0.19 m, y = +0.1 m. A third charge of +13.2 micro-Coulombs is placed at x = -0.19 m, y = 0 m. At what angle is the total electric filed is directed at the the point x = 0 m, y = +0.1 m if measured counter-clockwise from the positive x axis? Answer in degrees.

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A charge of +10.7 micro-Coulombs is placed at the origin of a coordinate system. Another charge of -12.6 micro-Coulombs is placed at x = +0.19 m, y = +0.1 m. A third charge of +13.2 micro-Coulombs is placed at x = -0.19 m, y = 0 m. At what angle is the total electric filed is directed at the point x = 0 m, y = +0.1 m if measured counter-clockwise from the positive x axis? Answer in degrees.

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Solution:

(please see the attached file for the figure)

y(m) E1 E3

0.1 P Î¸ E2 r2 Q2 = -12.6 Î¼C

r3 r1

Q3 = +13.2 Î¼C Î¸ Q1= +10.7Î¼C x (m)

-0.19 +0.19

By definition, an electric field at a point is defined as the force experienced by a unit +ve charge placed at that point. To determine the magnitude and direction of electric field at point P, we first determine the same at the given point due to each of the three charges and then determine their resultant (superposition principle).

Electric field at P due to Q1

E1 = kQ1/r12 where k = 1/4ÐŸÎµ0 = 9 x 109

Substituting Q1= +10.7Î¼C and r1 = 0.1 m

E1 = 9x109x10.7x10-6/0.12 = 9.63 x 106 N/C

Unit +ve charge placed at P will be repelled by the +ve charge Q1. Hence, direction of vector E1 is as shown in the fig..

Electric field at P due to Q2

E2 = kQ2/r22

Substituting Q2= -12.6Î¼C and r2 = 0.19 m

E2 = 9x109x12.6x10-6/0.192 = 3.14 x 106 N/C

Unit +ve charge placed at P will be attracted by the -ve charge Q2. Hence, direction of vector E2 is as shown in the fig..

Electric field at P due to Q3

E3 = kQ3/r32

Substituting Q3= +13.2Î¼C and r3 = âˆš(0.192 + 0.12) = 0.22 m

E3 = 9x109x13.2x10-6/0.222 = 2.45 x 106 N/C

Unit +ve charge placed at P will be repelled by the +ve charge Q3. Hence, direction of vector E3 is as shown in the fig..

From the fig., angle Î¸ = tan-1(0.1/0.19) = 27.76O

Resolving E3 along x and y axes we get: E3x = E3cosÎ¸ = 2.45x106xcos27.76O = 2.17 x 106 N/C

E3y = E3sinÎ¸ = 2.45x106xsin27.76O = 1.14 x 106 N/C

Net force along x axis = Ex = E2 + E3x = 3.14 x 106 + 2.17 x 106 = 5.31 x 106 N/C

Net force along y axis = Ey = E1 + E3y = 9.63 x 106 + 1.14 x 106 = 10.77 x 106 N/C

Magnitude of resultant electric field at P = E = âˆš[(5.31x106)2 + (10.77x106)2] = 12 x 106 N/C

Angle Î± made by vector E with x axis: Î± = tan-1(Ey/Ex) = tan-1(10.77 x 106/5.31 x 106) = 63.76O

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