Field of a point charge at 9 cm

Text

A point charge $Q = +5\;\mu\text{C}$ is placed in vacuum. Calculate the magnitude of the electric field generated at a distance $r = 9{,}0\;\text{cm}$ from the charge and state the direction and sense relative to the charge.

Solution

Formula. The magnitude of the electric field of a point charge at distance $r$ is $E = k\,\frac{|Q|}{r^2}$

Numerical substitution. With $r = 9{,}0\;\text{cm} = 0{,}090\;\text{m}$ and $r^2 = 8{,}1\cdot 10^{-3}\;\text{m}^2$: $E = (9\cdot 10^9)\,\frac{5\cdot 10^{-6}}{8{,}1\cdot 10^{-3}} = \frac{4{,}5\cdot 10^{4}}{8{,}1\cdot 10^{-3}}\;\text{V/m}$

Result. $\ev{E \approx 5{,}6\cdot 10^{6}\;\text{V/m}}$

Direction and sense. The field is radial: the lines emerge from the charge along the line joining charge and point. Since $Q > 0$, the sense is outward, i.e. directed from the charge towards the outside (moving away from the charge). A positive test charge placed at that point would be pushed away radially.

Links

Topics: Electric field and potential Concepts: Electric field · Coulomb’s law · Electric charge Objects: Charged sphere