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

The magnitude of the electric field on the y axis at y = 3.0 m is 23.1 N/C

How to determine the magnitude of the electric field on the y axis?

We have;

`OP = 3 m , OQ = 4 m`

`PQ = √(4^2+3^2) =5m`

`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)=27N/C`

Electric field at P due to the charge q is

`E_2=(Kq)/(PO^2) =(9 * 10^9 *8 * 10^-^9)/(9) =8N/C`

From 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 * 4 / 5 = - 21.6 N/C`

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

`Ey = 27 * 3 /5 - 8 = 8.2 N/C`

The resultant electric field at P is given by

`E=√(E^2_x + E^2_y) = √((-21.6)^2+(8.2)^2) = 23.1 N/C`

Therefore, the magnitude of the electric field on the y axis at y = 3.0 m is 23.1 N/C