Question

Three charges are at the corners of an equilateral triangle, as shown in the figure below. Calculate the electric field at a point midway between the two charges on the X-axis (Let q_1 = 3.50 mu C, q_2 = 5.00 mu C, and q_3 = 7.50 mu c.)

Answer 1

Solution 0.5m/0.5m 60° +02 G3 S.

To find the distance riis, opp sin θ =- sin 60°__i 0.5 = 0.5 sin 60° = 0.433 m The distance , and is. = 0.25m = 0.25m Since the electric field at the charge qis, (98109)(3.50×10-6) 0.433 - 168.0x10' N/C

Since the electric field at the charge q2is, ka2 (9x109)(5.0x10-5) (0.25) 720x103 NC Since the electric field at the charge q is, 9x10 (750x10-9) (0.25) 1080x10 NC Since the x component of the electric field is -720x103-1080x103 -360 x 103 NC

ponent of the electric field is, E,E -168 x10 NC The net electric field is, -397.3x10 N/C Hence magnitude of the net electric field is 397.3x103 N/C The direction of the electric field is o-tan-168 360 -25° Hence the direction of the electric field is 250below the +x axis