Three charges, q1 = 8.31nC, q2 = -5.92nC and q3 = 2.57nC are at the corners of a
ID: 1441494 • Letter: T
Question
Three charges, q1 = 8.31nC, q2 = -5.92nC and q3 = 2.57nC are at the corners of an equilateral triangle, as shown in the figure below.
The angle ? is 60.0o and L = 0.465 m. We are interested in the point midway between the charges q1 and q2 on the x axis.
For starters, calculate the magnitude of the electric field due only to charge q3 at this point.
Now calculate the magnitude of the electric field from all three charges at a point midway between the two charges on the x axis.
Calculate the angle of the electric field relative to the positive (to the right) x-axis, with positive values up (Counter-Clock Wise) and negative down (Clockwise).
Explanation / Answer
Electric field only due to q3 = kq3 / r2 =9×10^9 ×2.57×10^-9 /{(3/4)×0.465×0.465} = 142.6 N/C (- j)
Electric field due to q1 and q2
= 9×10^9 ×{8.31 - -5.92}×10^-9/{0.25 × 0.465×0.465}
=2369 N/C i
net Electric field due to all will be found by taking the resultant= 2373.5 N/C
Cos theta = 0.9981
theta = - 3.53 degrees
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