Question

Two point charges are located on the y-axis as follows: one charge q1 = -1.95 nC located at y= -0.550 m, and a second charge q2 = 3.60 nC at the origin (y=0). What is the magnitude of the total force exerted by these two charges on a third charge q3 = 4.90 nC located at y3 = -0.435 m ?

Answer 1

Given :

Three charges :

`q_1=-1.95\ nC\\\\q_2=3.60\ nC\\\\q_3=4.90\ nC`

There location on y-axis :

`y_1=-0.550\ m\\\\y_2=0\ m\\\\y_3=-0.435\ m`

To Find :

Total force acting on
`q_3` .

Solution :

Total force acting on
`q_3` is :

`F_3=(kq_1)/(r_1^2)+(kq_2)/(r_2^2)\\\\F_3=k[((1.95* 10^(-9))/(0.550-0.435))+((3.6* 10^(-9))/(0.550-0))]\\\\F_3=9.0 * 10^9 * [((1.95* 10^(-9))/(0.550-0.435))+((3.6* 10^(-9))/(0.550-0))]\\\\F_3=211.52\ N`

Therefore , force applied on
`q_3` is 211.52 N .