Question

Open-Response Homework Problem 3.1 A +2.0nC charge is located at (-1.5m, 0). An -2.0nC charge is located at (+1.5m,0. The point P is located at (0, 2.6m). (a)[4 pt (s)_]Draw the above system, and draw the individual and resultant electric field vectors at point P. (b)[6 pt(s)_Find the electric field at point P. (c)[2 pt(s) IA-10μC charge is placed at point P, find the electric force on this charge.

Answer 1

(a) Let q_1 = + 2 n C q_2 = -2 n C q_2 = -q_1 rightdoublearrow Vertical-bar q_2 Vertical-bar = Vertical-bar q_1 Vertical-bar Let angle b/w AP and AO is theta angle b/w PB and BO is theta BP = AP = r squareroot (1.5)^2 + (2.6)^2 E_1 rightarrow = electric field vector at P due to charge (+ 2n () E_2 rightarrow = electronic field vector at P due to charge (- 2n () E rightarrow Resultant electronic field vector a P (b) electronic field at P due to charge q_1 E_1 = q_1/4 pi epsilonvariant_0 r^2 electronic field at P due to charge q_2 E_2 = q_2/4 pi epsilonvariant_0 r^2 E-2 = -q_1/4 pi epsilonvariant_0 r^2 E_1 sin theta and E_2 sin theta are equal in magnitude and opposite in direction . so cancel each other E = E_1 cos theta + E_2 cos theta = q_1/4 pi epsilonvariant_0 r^2 cos theta cos theta = 1.5/r E = q_1/4 pi epsilonvariant_0 r^3 times 1.5 =