at what distance from a 0.100 c charge will the electric field be 10000 n/c? (unit = m)

Explanation:

Step1: Recall electric - field formula

The electric - field formula is $E = \frac{kQ}{r^{2}}$, where $E$ is the electric field, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$ is the Coulomb's constant, $Q$ is the charge, and $r$ is the distance from the charge. We need to solve for $r$.

Step2: Rearrange the formula for $r$

Starting from $E=\frac{kQ}{r^{2}}$, we can cross - multiply to get $Er^{2}=kQ$, then $r^{2}=\frac{kQ}{E}$, and $r = \sqrt{\frac{kQ}{E}}$.

Step3: Substitute the given values

Given $Q = 0.100\ C$, $E = 10000\ N/C$, and $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$. Substitute these values into the formula: $r=\sqrt{\frac{9\times10^{9}\ N\cdot m^{2}/C^{2}\times0.100\ C}{10000\ N/C}}$. First, calculate the value inside the square root: $\frac{9\times10^{9}\times0.100}{10000}=\frac{9\times10^{8}}{10^{4}} = 9\times10^{4}$. Then, find the square root: $r=\sqrt{9\times10^{4}\ m^{2}}=300\ m$.

Answer:

$300$