at what distance from a 2.45*10^-6 c charge will the electric field be 100 n/c? (unit = m)

Explanation:

Step1: Recall electric - field formula

The electric - field formula for a point charge is $E = k\frac{q}{r^{2}}$, where $E$ is the electric field, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$ is 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 $E = k\frac{q}{r^{2}}$, we can cross - multiply to get $Er^{2}=kq$, and then $r^{2}=\frac{kq}{E}$. Taking the square root of both sides, we have $r=\sqrt{\frac{kq}{E}}$.

Step3: Substitute the given values

Given $q = 2.45\times10^{-6}\ C$, $E = 100\ N/C$, and $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$. $r=\sqrt{\frac{9\times 10^{9}\times2.45\times10^{-6}}{100}}$ First, calculate the value inside the square root: $9\times 10^{9}\times2.45\times10^{-6}=9\times2.45\times10^{9 - 6}=22.05\times10^{3}=22050$. Then, $\frac{22050}{100}=220.5$. So, $r=\sqrt{220.5}\approx14.85\ m$.

Answer:

$14.85$