Certainly! Let's solve the problem step by step.
Given:
- The charge [tex]\( q \)[/tex] is [tex]\( 6.93 \times 10^{-4} \)[/tex] Coulombs.
- The potential energy [tex]\( U \)[/tex] is [tex]\(-3.09\)[/tex] Joules.
We need to find the electric potential [tex]\( V \)[/tex] at that point.
The electric potential [tex]\( V \)[/tex] is defined as the potential energy per unit charge. Mathematically, it can be expressed as:
[tex]\[ V = \frac{U}{q} \][/tex]
Now, let's substitute the given values into this formula:
- [tex]\( U = -3.09 \)[/tex] Joules
- [tex]\( q = 6.93 \times 10^{-4} \)[/tex] Coulombs
Therefore,
[tex]\[ V = \frac{-3.09 \text{ J}}{6.93 \times 10^{-4} \text{ C}} \][/tex]
When you divide the potential energy by the charge, you get:
[tex]\[ V = -4458.8744588744585 \text{ V} \][/tex]
So the electric potential [tex]\( V \)[/tex] at that point, including the sign, is:
[tex]\[ V = -4458.87 \text{ V} \][/tex]
This means the electric potential at that point in space is [tex]\(-4458.87\)[/tex] volts.
