Certainly! Let's find the distance from a charge of [tex]\(1.00 \times 10^{-5}\)[/tex] Coulombs where the electric potential is 10000 Volts.
First, let's recall the formula for electric potential [tex]\(V\)[/tex] at a distance [tex]\(r\)[/tex] from a point charge [tex]\(q\)[/tex]:
[tex]\[ V = \frac{k \cdot q}{r} \][/tex]
where:
- [tex]\(V\)[/tex] is the electric potential in Volts (V),
- [tex]\(k\)[/tex] is Coulomb's constant, approximately [tex]\(8.99 \times 10^{9} \, \text{N m}^2/\text{C}^2\)[/tex],
- [tex]\(q\)[/tex] is the charge in Coulombs (C),
- [tex]\(r\)[/tex] is the distance from the charge in meters (m).
We need to solve for [tex]\(r\)[/tex]. Let's rearrange the formula to isolate [tex]\(r\)[/tex]:
[tex]\[ r = \frac{k \cdot q}{V} \][/tex]
Now, we can substitute the given values into the equation:
- [tex]\(q = 1.00 \times 10^{-5} \, \text{C}\)[/tex]
- [tex]\(V = 10000 \, \text{V}\)[/tex]
- [tex]\(k = 8.99 \times 10^{9} \, \text{N m}^2/\text{C}^2\)[/tex]
Substituting these values:
[tex]\[ r = \frac{8.99 \times 10^{9} \, \text{N m}^2/\text{C}^2 \times 1.00 \times 10^{-5} \, \text{C}}{10000 \, \text{V}} \][/tex]
Performing the multiplication in the numerator:
[tex]\[ 8.99 \times 10^{9} \times 1.00 \times 10^{-5} = 8.99 \times 10^{4} \][/tex]
Next, divide by 10000 V:
[tex]\[ r = \frac{8.99 \times 10^{4}}{10000} \][/tex]
[tex]\[ r = 8.99 \, \text{m} \][/tex]
Therefore, the distance from a [tex]\(1.00 \times 10^{-5} \, \text{C}\)[/tex] charge where the electric potential is 10000 V is accurately calculated to be:
[tex]\[ r \approx 8.99 \, \text{m} \][/tex]
