Example 18.2 from College Physics, 18.4 Electric Field: Concept of a Field Revisited

Calculating the Electric Field of a Point Charge

Calculate the strength and direction of the electric field $E$ due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
We can find the electric field created by a point charge by using the equation $E = kQ / r^{2}$ .
Solution
Here $Q = 2 . 00 \times 10^{- 9}$ C and $r = 5 . 00 \times 10^{- 3}$ m. Entering those values into the above equation gives

$\begin{array}{l} E & = & k \frac{Q}{r^{2}} \\ = & (8.99 \times 10^{9} N \cdot m^{2} {/C}^{2}) \times \frac{(2.00 \times 10^{- 9} C)}{(5.00 \times 10^{- 3} m)^{2}} \\ = & 7.19 \times 10^{5} N/C. \end{array}$
18.14
Discussion
This electric field strength is the same at any point 5.00 mm away from the charge $Q$ that creates the field. It is positive, meaning that it has a direction pointing away from the charge $Q$ .

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