Question

e) Find the electric Field vector at point (0, +a) (Calculate the magnitude, and draw the vector.) f Sketch the electric field lines. g) Find the electric Field at (x, 0) for「a. h) Find out the location (x, y) where the electric Field E becomes zero. (Hint: Use the solution of e).) +4Q -Q

Answer 1

E_1 = k (4d)/(squareroot 2a)^2 = 4 kQ/a^2 = 2 kQ/a^2 E_2 = k (Q)/squareroot 2a^2 = kQ/2a^2 E_net = (E_1 sin 30 - E_2 sin 30) I + (E_1 + E_2) cos 30j = (kQ/a^2 - kQ/4 a^2) i + 5/2 v_3/2 kQ/a^2 j E_net = 3 kQ/4 a^2 I + 5 squareroot 3 kQ/4 a^2 I Vertical-bar E_net Vertical-bar = squareroot (3/4 kQ/a^2)^2 + (5 squareroot 3 k Q/4 a^2)^2 Vertical-bar E_net Vertical-bar = 2.2912 kQ/a^2 Electric field lines let wear E = kQ/(x - a)^2 + 4 kQ/(x + 2a)^2 kQ/(x - a)^2 = -4 kQ/(x + 2a)^2 E = 0 kQ/(x - a)^2 = -4 kQ/(x + 2a)^2 (x + 2a)^2 = -4 (x - a)^2 x^2 + 4a^2 + 4a x = -4 (x^2 + a^2 - 2a x) x^2 + 4a^2 + 4a x = -4 x^2 - 4a^2 + 8a x 5x^2 + 8a^2 - 4a x = 0 solving quadratic equation x = 4a plusminus squareroot (4a) - 4 times 5 times 8a^2/2 times 5 x =