Question

12. 2 are -1.52 C and +3.20 C respectively. -12 points What is the electric field (magnitude and direction) at point in the figure below due to the two charges? Charges 01 and magnitude N/C direction counterclockwise from the +x-axis 6.83 x 10m 30.0 3.19 x 10 m

Answer 1

Summary-This is the basic problem from Electrostatic. I have provided the step by step solution.

Sol Let &, and Ez a q2 = 3.2 uc Let E and Ea are the Electric fields at due to 2, and a respectively. K21 - 9x10x1.52x10-6 D (3-19X16-6) = 2.344 x 1015 N/ 30°r, 9, 2 ,- x 3 V $79x16-02 Ezy E- Kq2 3 x 10 x 3.2x106 9x10°x32x104 (6.83 x 1062 = 0.617X10'S NIC Resolving Ez along x and yaxes. Eax = Ez Cos300 = 0.617x105x0.866= 0.534 x 105 N/C Egy = E₂ sin 30 = 0.617x105x0.5 = 0.3085 X10SN/C Net electric field along x-axis, Ex= E,-E2x = 1-3 44x10's - 0.534x10's = 0.81XIO'S N/c alony y-axis = Ey = - Ezy = – 0.3085 810' N/C Net Electric field, Ek J E² + Ey? J(0.81x10'S)*+10:3085 X 105 ) 2 = 0.866 * 1015 N/C direction of E = (360-0) Counterclockwise from x-axis. aut 10.308531015) 90.85 o = tant (E = tant (0 300 st hence, direction = 360-20.85 = 339.15 counter clock wise form x-axis. (0.81*10' 5 00