Question

What is the electric field at a position that is 1.6m East of a point charge of 7.2 x 10^-6C?

Answer 1

The electric field produced by a single point charge is given by:

`E(r) = k (q)/(r^2)`
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge

In our problem, the charge is
`q=7.2 \cdot 10^(-6) C`, while the distance is
`r=1.6 m`, therefore the intensity of the electric field at that distance is

`E=k (q)/(r^2)=(8.99 \cdot 10^(9) Nm^2C^(-2)) ((7.2 \cdot 10^(-6) C))/((1.6m)^2) =2.53 \cdot 10^4 N/C`
and since the charge is positive, the electric field is directed away from the charge (so, eastwards)