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

Answer:

$E = \frac{8.99 x10^{9} \frac{Nm^2}{C^2} * 2x10^{-9} C}{(0.64m)^2} =43.85N$

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:

$E = \frac{K Q}{d^2}$

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

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

$d =\sqrt{0.41}= 0.64m$

And now we can replace into the formula since we know that $Q = 2nC= 2x10^{-9}C$ and $K = 8.99 x10^{9} \frac{Nm^2}{C^2}$, and we got:

$E = \frac{8.99 x10^{9} \frac{Nm^2}{C^2} * 2x10^{-9} C}{(0.64m)^2} =43.85N$