Question

Three charges are placed on the corners of an equilateral triangle (10μC at the top, -10μC at the lower right, and 5μC at the lower left). Find the magnitude and the direction of the electric field at the center (Point P) of the triangle. The length of one of the sides is 0.5 m.

(a) 28.8 x 106 N/C, θ = 36o

(b) 1.93 x 106 N/C, θ = -44o

(c) 1.02 x 107 N/C, θ = -57o

(d) zero

I'd really appreciate the help!

Answer 1

43 30 ら

r = distance of each corner from the center = L/(2 sqrt(3)) = 0.5 /(2 sqrt(3)) = 0.14 m

E_a = k q_a/r² = (9 x 10⁹) (10 x 10^-6)/(0.14)² = 4.6 x 10⁶ N/c

E_b = k q_b/r² = (9 x 10⁹) (5 x 10^-6)/(0.14)² = 2.3 x 10⁶ N/c

E_c = k q_c/r² = (9 x 10⁹) (10 x 10^-6)/(0.14)² = 4.6 x 10⁶ N/c

Net electric field along the X-direction is given as

E_x = E_c Cos30 + E_b Cos30 = (4.6 x 10⁶ ) Cos30 + (2.3 x 10⁶ ) cos30 = 5.98 x 10⁶ N/C

Net electric field along the Y-direction is given as

E_y = - E_a - E_c Cos60 + E_b Sin60 = - (4.6 x 10⁶ ) - (4.6 x 10⁶ ) Cos60 + (2.3 x 10⁶ ) sin30 = - 5.75 x 10⁶ N/C

net electric field is given as

E = sqrt(E_x² + E_y²) = sqrt((5.98 x 10⁶ )² + ( - 5.75 x 10⁶ )²) = 8.3 x 10⁶ N/C

direction :

tan^-1(E_y/E_x) = tan^-1(5.75 x 10⁶ /(5.98 x 10⁶ )) = - 44