Two point charges of the same magnitude but opposite signs are fixed to either e
ID: 2233598 • Letter: T
Question
Two point charges of the same magnitude but opposite signs are fixed to either end of the base of an isosceles triangle, as the drawing shows. The electric field at the midpoint M between the charges has a magnitude EM. The field directly above the midpoint at point P has a magnitude EP. The ratio of these two field magnitudes is EM/EP = 7.76. Find the angle ? in the drawing. If there is a cos^3 in the problem please explain how to use that as well. I understand the problem except how to solve with the cos^3 in it and thats the only answer I have found. I need a complete answer!!!!!!!
Explanation / Answer
Sorry, correct answer:
EM = 2kq/(L/2)^2 = 8 kq/L^2
EP = 2kq/(L/2cos(a))^2 (cos(a)) = 8 kq/L^2 * (cos(a))^3
==> EM/EP= 1/(cos(a))^3 = 7.76
==> cos(a)^3 = 1/7.76 = 0.12887
==> cos(a) = (0.12887)^(1/3)
==> cos(a) = 0.5051
==> a = 59.7 degrees
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