a source charge of 5.0 μc generates an electric field of 3.93×10^5 \frac{n}{c} at the location of a test…

Explanation:

Step1: Recall electric - field formula

The electric - field formula is $E = k\frac{q}{r^{2}}$, where $E$ is the electric field, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q$ is the source charge, and $r$ is the distance from the source charge to the test charge. We need to solve for $r$. First, re - arrange the formula for $r$: $r^{2}=k\frac{q}{E}$, so $r=\sqrt{\frac{kq}{E}}$.

Step2: Convert source - charge unit

The source charge $q = 5.0\ \mu C=5.0\times10^{- 6}\ C$, and $E = 3.93\times10^{5}\ N/C$, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$. Substitute the values into the formula: $r=\sqrt{\frac{9\times10^{9}\times5.0\times10^{-6}}{3.93\times10^{5}}}$.

Step3: Calculate the value of $r$

First, calculate the value inside the square - root: $\frac{9\times10^{9}\times5.0\times10^{-6}}{3.93\times10^{5}}=\frac{45\times10^{3}}{3.93\times10^{5}}=\frac{45}{3.93\times10^{2}}\approx0.1145$. Then, $r=\sqrt{0.1145}\approx0.34\ m$.

Answer:

0.34 m