Three charges are at the corners of an equilateral triangle, as shown in the fig

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

q1 = 6.50 ?C

q2 = 2.50 ?C,

q3 = ?4.50 ?C.)

I solved the Electric field for each q, and calculated the magnitude but I'm not sure if I did it correctly. In additon, I don't know how to find direction for three charges.

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 6.50 pc, q2 2.50 ALC, and 93 4.50 uc.) N/CE magnitude below the +x-axis direction 0.500 m 60.00

Explanation / Answer

q1 = 6.5 mu C ; q2 = 2.5 mu C and q3 = -4.5 mu C

d = 0.5 m

The field at a point midway between the two charges on the x-axis can be calculated as follows:

Field due to q1, q2 and q3 will be:

E1 = k q1/r1^2 ; E2 = k q2/r2^2 ; E3 = k q3/ r3^2

r1 = 0.5 x sin60 = 0.433 ; r2 = 0.25 m ; r3 = 0.25

E1 = 9 x 10^9 x 6.5 x 10-6 / 0.433^2 = 3.12 x 10^5 N/C

E2 = 9 x 10^9 x 2.5 x 10^-6 / 0.25^2 = 3.6 x 10^5 N/C

E3 = 9 x 10^9 x 4.5 x 10^-6/0.25^2 = 6.5 x 10^5 N/C

We see that the field due to 1 will be in y direction and that due to 2 and 3 are in x direction, so the total field in x direction is:

Ex = E2 + E3 = 10^5 (3.6 + 6.5) = 10.1 x 10^5 N/C

Ey = 3.12 x 10^5 N/C

E = sqrt [(3.12 x 10^5)^2 + (10.1 x 10^5)^2] = 1.06 x 10^6 N/C

direction:

theta = tan -1 (3.12/10.1) = 17.17 deg

Hence, E = 1.06 x 10^6 N/C and direction = 17.17 deg

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