Question

The Electric Field 9.0m from a +25μC charge is

Answer 1

The electric field intensity generated 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 this problem, the charge is
`q=25 \mu C=25 \cdot 10^(-6) C` and we are asked to calculate the field at distance
`r=9.0 m`, so the electric field is

`E(9.0 m)=(8.99 \cdot 10^9 Nm^2C^(-2))(25 \cdot 10^(-6) C)/((9 m)^2)=2775 V/m`