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

$E(r)=k\frac{q}{r^2}$

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

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