Question

Three point charges are arranged as shown in the figure below. 

(a) Find the vector electric field that the q = 7.00 nC and -3.00 nC charges together create at the origin.

(b) Find the vector force on the 5.00 nC charge.

Answer 1

Part a. Then electric field due to 7.0 nC charge is

E1 = kq/R^2 = 9*10^9 *7.0*10^-9* /0.3*0.3

E1= -700 i N/C

EF due to -3 nC charge is

E2 = 9*10^9 * -3*10^-9/0.1*0.1 = -2700 j N/C

Etotal = (-700 i )+ (- 2700 j) N/C

Part b. electric force F = Eq

Then put values

=-700*5*10^(-9) i + (-2700)*5*10^(-9) j

F = -3.5 μN i - 13.5 μN j (Answer)

I hope help you !!