at what distance from a -5.35*10^-6 c charge will the electric potential be -500 v? (unit = m)

Explanation:

Step1: Recall electric - potential formula

The formula for electric potential due to a point charge is $V = \frac{kQ}{r}$, where $V$ is the electric potential, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $Q$ is the charge, and $r$ is the distance from the charge. We need to solve for $r$.

Step2: Rearrange the formula for $r$

From $V=\frac{kQ}{r}$, we can get $r=\frac{kQ}{V}$.

Step3: Substitute the given values

We have $Q=- 5.35\times10^{-6}\ C$, $V = - 500\ V$, and $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$. $r=\frac{9\times10^{9}\times(-5.35\times10^{-6})}{-500}$ $r=\frac{-48150}{-500}=96.3\ m$

Answer:

$96.3$