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Answer on Electromagnetism Question for Kamilla

Question #1622
A charge of 3.00 x 10-6 C is located at the point x = + 0.100 m, y = 0.0. A second charge of -6.00 x 10-6 C is located at x = -0.100 m, y=0.0. (a) What are the x and y components of the electric field at the point x = 0.0, y = + 0.300 m? (b) What are the magnitude and direction of the electric field at that same point?
Expert's answer
The distance from the point (0.1,0) to (0,0.3) is R = √0.1, from the point (-0.1, 0) to (0, 0.3) it's the same of √0.1.
The magnitude of the electric field of the first charge is
E1 = (1/4π ε0)q1/R = - 8.99x109*3x10-6/√0.1 = - 85.29x103 [V/m]
The magnitude of the electric field of the second charge is
E2 = (1/4π ε0)q2/R = 8.99x109*6x10-6/√0.1 = 170.57x103 [V/m]
The x component of the total electric field is Ex = - 0.1/√0.1( 85.29 +170.57)x103 = - 80.91x103 [V/m]
The y component is Ey = 0.3/√0.1( 85.29-170.57)x103 = -80.91x103 [V/m]
The magnitude of the total electric field would be E2 = (Ex2+Ey2) = 2* 80.91x106 , E = 114.42 x103 [V/m], it's directed along the angle of 225 grad from the positive x-axis (tan(θ) = Ey/Ex = 1).

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