Question

Three Charges are located at the corners of an equilateral triangle as show below

A) Calculate the electric field, E, in magnitude and direction, at the point midway between the 8.00nC and -5.00nC charges on the right hand side of the triangle.

B) If a charge q = -6.00nC is placed at this midpoint location, what force F, in magnitude and direction would it experience?

C) If the charge in part B was only made of electrons (no protons at all), what would the initial acceleration, in magnitude and direction, of this charge be if it were released?

(please see below for a screen shot of the triangle, the distance between points is .750 m)

Snt sdo sokan ひ109) E = r2 a 12.0 nC S.169) 10 0.750 m 60 40 8.00 nC -5.00 nC

Answer 1

Given that so we know that F = K q_1 q_2/r^2 and E = K q/r^2 so E_1 = K q/r^2 d = squareroot (1.5)^2 + (0.75)^2 d = 1.677 nm so E_1 = 9 times 10^9 times 12 times 10^3/(1.677)^2 E_1 = 108/2.8125 = 38.4 V/m E_1 = 38.4 V/m (- y) E_2 = 9 times 10^9 times 8 times 10^- 9/(0.75)^2 = 72/0.5625 E_2 = 1 = 8 V/m (x) E_3 = 9 times 10^9 times 5 times 10^- 9/(0.75)^2 = 45/0.5625 E_3 = 80 V/m x so E_net = E_1 + E_2 + E_3 E_net = 80 x + 128 x - 38.4 y E_net = 208 x - 38.4 y (ii) we know that F = q B given q = -6 n C so F = -6