Positive point charges are placed at three corners of a rect

Positive point charges are placed at three corners of a rectangle, as shown in the figure, in which d = 0.43 m. Take the +x-axis to point to the right. (a) What is the electric field at the fourth corner? N/C at °, measured counterclockwise from the -x-axis (b) A small object with a charge of +7.6 µC is placed at the fourth corner. What force acts on the object? N at °, measured counterclockwise from the -x-axis

Solution

Electric Field due to the left negative charge is,

E1 = 9*10^9*1*10^-6/0.2^2 = 225*10^3 N/C = 225000 N/C

Electric Field due to right up positive charge is,

   E2 = 9*10^9*4*10^-6/0.43^2 = 194699.84 N/C = 194.7*10^3 N/C

diagonal distance = sqrt(0.2^2+0.43^2) = 0.4742 m

Electric Field due to right positive charge is,

E3 = 9*10^9*6*10^-6/0.4742^2 = 240143.4 N/C = 240.143*10^3 N/C

x-components:

E3(x-axis) = E3*cos = 240.143*10^3*0.43/0.4742 = 217759.73 N/C

E3(y-axis) = E3*sin = 240.143*10^3*0.2/0.4742 = 101283.42 N/C

Thus, the net electric field,

Enet(x-axis) = 194699.84 + 217759.73 = 412459.57 N/C

Enet(y-axis) = -225000+101283.42 = -123716.58 N/C

the net electric field is,

E(net) = sqrt(412459.57^2+(-123716.58)^2)*10^3 = 430614.3 N/C

Angle = -16.7^o, from -x-axis

(a)

E(net) = 430614.3 N/C, at Angle = 16.7^o

(b)

the net force is,

    F(net) = (430614.3 N/C)*(7.6*10^-6 C) = 3.27 N at angle = 16.7^o