(A) For starters, calculate the magnitude of the electric field due only to char
ID: 1432280 • Letter: #
Question
(A) For starters, calculate the magnitude of the electric field due only to charge q3 at this point.
(B) Now calculate the magnitude of the electric field from all three charges at a point midway between the two charges on the x axis.
(c) 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). (enter the answer with units of deg)
(D) If a tiny particle with a charge q = 1.22nC were placed at this point midway between q1 and q2, what is the magnitude of the force it would feel?
Explanation / Answer
0.433^2 - (0.433/2)^2
(a)
E = k*q3/r^2
E = (8.9*2.71)/(L*cos(30))^2
E = (8.9*2.71)/(0.433*cos(30))^2
E = 171.5 N/C
(b)
Field due to q1 & q2,
E = k*q1/r^2 + k*q2/r^2
E = (8.9)/(0.433/2)^2 * [7.32 + 5.31]
E = 2398.2 N/C
Net Electric Field, E = sqrt(171.5^2 +2398.2^2 )
E = 2404.3 N/C
(c)
Direction = tan^-1(171.5/2398.2)
Direction = - 4.1o [Clockwise from +ve x axis.]
(d)
F= q*E
F = 1.22 * 10^-9 * 2404.3 N
F = 2.93 * 10^-6 N
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