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

d = squareroot a^2 + a^2 = squareroot 2 a Vertical-bar From the figure we can get, phi = 45 degree Therefore angle between E^+ & E^- Therefore Vertical-bar E^+ Vertical-bar = 1/4 pi elementof_0 4Q/(squareroot 2 a)^2 Vertical-bar E^- Vertical-bar = + 1/4 pi elementof_0 Vertical-bar -Q Vertical-bar /(squareroot 2 a)^2 = 1/4 pi elementof_0 Q/(squareroot 2 a)^2 Therefore E Net electric field E = squareroot Vertical-bar E^+ Vertical-bar ^2 + Vertical-bar E^- Vertical-bar ^2 + 2 Vertical-bar E^+ Vertical-bar Vertical-bar E^- Vertical-bar cos 90 degree = squareroot (1/4 pi elementof_0 4Q/2a^2)^2 + (1/4 pi elementof_0 Q/2a^2)^2 \r\n = 1/4 pi elementof_0 Q/2a^2 squareroot 4^2 + 1^2 E = 1/4 pi elementof_0 Q/2a^2 squareroot 17 Now, tan theta = Vertical-bar E^- Vertical-bar / Vertical-bar E^+ Vertical-bar = 1/4 Therefore theta = tan^-1 (1/4) = 14 degree Therefore E vector at point p is of magnitude 1/4 pi elementof_0 Q/2a^2 squareroot 17 and directed towards positive x - axis making an angle (45 degree + c) degree = 59 degree with the y - axis. \r\n Therefore Electric field