# Question #12904

Oct 10, 2017

$d = 2$ m

#### Explanation:

$F = k \frac{q \cdot Q}{d} ^ 2$

and

$E = \frac{F}{d}$

Plugging $F$ into our $E$ equation, we get:

$E = \frac{F}{q} = \frac{\frac{k \cdot \cancel{q} \cdot Q}{d} ^ 2}{\cancel{q}} = \frac{k \cdot Q}{d} ^ 2$

Where $k$ is Coulomb's law constant: $8.99 \cdot {10}^{9} \frac{N \cdot {m}^{2}}{C} ^ 2$
...or more simply: $9 \cdot {10}^{9} \frac{N \cdot {m}^{2}}{C} ^ 2$

Convert to the appropriate units:

$\mu C \implies C$ and $\frac{k N}{C} \implies \frac{N}{C}$

$E = \frac{k \cdot Q}{d} ^ 2 \implies \left(2.5 X {10}^{3} \frac{N}{C}\right) = \frac{\left(9 \cdot {10}^{9} \frac{N \cdot {m}^{2}}{C} ^ 2\right) \left(1 \cdot {10}^{-} 6 C\right)}{d} ^ 2$

Solve for $d$:

$d = \sqrt{\frac{\left(9 \cdot {10}^{9} \frac{N \cdot {m}^{2}}{C} ^ 2\right) \left(1 \cdot {10}^{-} 6 C\right)}{2.5 X {10}^{3} \frac{N}{C}}} = 1.89 m$

$= \underline{2}$ meters for significant figures ($\underline{1} \mu C$)