Question

A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, what is the magnitude of the electric field on the y axis at y = +3.0 m?

Answer 1

23.1 N/C

Step-by-step explanation:

OP = 3 m , OQ = 4 m

`PQ = \sqrt{4^(2)+3^(2)}=5 m`

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

`E_(1)=(KQ)/(PQ^(2))=(9* 10^(9)* 75* 10^(-9))/(25)=27 N/C`

Electric field at P due to the charge q is

`E_(2)=(Kq)/(PO^(2))=(9* 10^(9)* 8* 10^(-9))/(9)=8 N/C`

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

`E=\sqrt{E_(x)^(2)+E_(y)^(2)}=\sqrt{(-21.6)^(2)+(8.2)^(2)}=23.1 N/C`