What is the magnitude of the electric field strength at a distance of 2.0 m from a point charge Q = 4.0 C? A) 54.0×10^9 N/C B) 18.0×10^9 N/C C) 9.0×10^9 N/C D) 72.0×10^9 N/C

What you are being asked

You need the electric field magnitude produced by a single point charge at a given distance. For a point charge, the field gets weaker with the square of the distance.

Using Coulomb’s law for electric field

For a point charge $Q$, the electric field magnitude a distance $r$ away is

$$E = k\frac{|Q|}{r^2}$$

where $k = 9.0\times 10^9\ \text{N·m}^2/\text{C}^2$.

Substitute the numbers

Given $Q = 4.0\ \text{C}$ and $r = 2.0\ \text{m}$:

$$E = (9.0\times 10^9)\frac{4.0}{(2.0)^2}$$

Compute the denominator:

$$(2.0)^2 = 4.0$$

So,

$$E = (9.0\times 10^9)\frac{4.0}{4.0} = 9.0\times 10^9\ \text{N/C}$$

Matching to the multiple-choice options

That value matches option C) $9.0\times 10^9\ \text{N/C}$.