Drill – distance for a given field

Drill: field of a point charge. At what distance from a point charge $q = 5\ \mu\text{C}$ is the magnitude of the electric field $E = 1000\ \text{N/C}$? Use $k = 9\cdot 10^9 \ \text{N m}^2/\text{C}^2$.

Solution

Setup. The magnitude of the field of a point charge is $E = \dfrac{kq}{r^2}$. Solving for the distance: $r = \sqrt{\frac{kq}{E}}$ Data. $q = 5\ \mu\text{C} = 5\cdot 10^{-6}$ C, $E = 1000$ N/C, $k = 9\cdot 10^9 \ \text{N m}^2/\text{C}^2$. Substitution. $r = \sqrt{\frac{(9\cdot 10^9)(5\cdot 10^{-6})}{1000}} = \sqrt{\frac{4{,}5\cdot 10^4}{10^3}} = \sqrt{45}$ Result. $\ev{r \approx 6{,}7 \ \text{m}}$

Links

Topics: Electric field and potential Concepts: Electric field · Coulomb’s law Skills: Superposition principle Methods: Gradient method Objects: Charged sphere