Question

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

Answer 1

Step-by-step explanation:

The electric potential can be calculated as:

`V=k(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*10^9)(-5.35*10^(-6))/(r)`

Then, solving for r:

`\begin{gathered} -500=(-48.15*10^3)/(r) \\ 500r=48.15*10^3 \\ r=(48.15*10^3)/(500) \\ r=96.3\text{ m} \end{gathered}`

Therefore, the distance is 96.3 m