Question

At what distance from a-5.35*10^-6 C charge will theelectric potential be -500 V?(Unit = m)IEnter

Answer 1

Explanation:

The electric potential can be calculated as:

$V=k\frac{Q}{r}$

Where k is 9 x 10⁹ N m²/C², r is the distance and Q is the charge.

So, replacing each value, we get:

$-500=(9\times10^9)\frac{-5.35\times10^{-6}}{r}$

Then, solving for r:

$\begin{gathered} -500=\frac{-48.15\times10^3}{r} \\ 500r=48.15\times10^3 \\ r=\frac{48.15\times10^3}{500} \\ r=96.3\text{ m} \end{gathered}$

Therefore, the distance is 96.3 m