Question

Two charges are arranged as shown in Figure Q1.1. Charge q, -48 pC is located on the positive y-axis at y 4.0 cm. Charge q2-100 pC is located on the positive x-axis atx-5.0 cm. 4.0 cm 5.0 cm 100 pc

Answer 1

5) electric field due to q1 = kq1/r^2

= 9*10^9*48*10^-12 / 0.04^2

= 270 N/C (in + ve y-direction)

electric field due to q2 = kq2 / r^2

= 9*10^9*100*10^-12/ 0.05^2

= 360 N/C (in +ve x-direction)

net EF = sqrt(270^2 + 360^2) = 450 N/C

b) direction = tan^-1 ( 360/270) = 53.13 ^o above the x-axis in counter clock wise direction

c) magnitude of electric force = qE = 20*10^-12*450 = 9*10^-9 N

d) direction will same as that of electric field = 53.13 ^o above the x-axis in counter clock wise direction