Question

What is the electric field at a location b → = <-0.5, -0.4, 0> m, due to a particle with charge +2 nC located at the origin?

Answer 1

`E = (8.99 x10^(9) (Nm^2)/(C^2) * 2x10^(-9) C)/((0.64m)^2) =43.85N`

Step-by-step explanation:

For this case we assume that we want to find the electrical field at the point P as we can see on the figure attached.

The electrical field wormula is given by:

`E = (K Q)/(d^2)`

Where r is the distance from the point and the charge. On this case we can use the Pythagoras theorem and we got:

`d^2 = (-0.5m)^2 +(-0.4)^2 = 0.41m^2`

`d =√(0.41)= 0.64m`

And now we can replace into the formula since we know that
`Q = 2nC= 2x10^(-9)C` and
`K = 8.99 x10^(9) (Nm^2)/(C^2)`, and we got:

`E = (8.99 x10^(9) (Nm^2)/(C^2) * 2x10^(-9) C)/((0.64m)^2) =43.85N`