Get the most out of your questions with the extensive resources available on IDNLearn.com. Join our Q&A platform to receive prompt and accurate responses from knowledgeable professionals in various fields.
Sagot :
To determine the distance from a [tex]\(-5.35 \times 10^{-6}\)[/tex] C charge where the electric potential is [tex]\(-500\)[/tex] V, we can use the formula for electric potential due to a point charge:
[tex]\[ V = \frac{k \cdot q}{r} \][/tex]
where:
- [tex]\(V\)[/tex] is the electric potential,
- [tex]\(k\)[/tex] is Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2}\)[/tex]),
- [tex]\(q\)[/tex] is the charge,
- [tex]\(r\)[/tex] is the distance from the charge.
Given:
- [tex]\(V = -500 \, \text{V}\)[/tex],
- [tex]\(q = -5.35 \times 10^{-6} \, \text{C}\)[/tex],
- [tex]\(k = 8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2}\)[/tex].
We need to solve for [tex]\(r\)[/tex]:
[tex]\[ r = \frac{k \cdot q}{V} \][/tex]
Plugging in the given values:
[tex]\[ r = \frac{8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \cdot (-5.35 \times 10^{-6} \, \text{C})}{-500 \, \text{V}} \][/tex]
Simplify the expression:
[tex]\[ r = \frac{(8.99 \times 10^9) \times (-5.35 \times 10^{-6})}{-500} \][/tex]
[tex]\[ r = \frac{-4.8065 \times 10^4}{-500} \][/tex]
[tex]\[ r = 96.193 \, \text{m} \][/tex]
Thus, the distance from the [tex]\(-5.35 \times 10^{-6}\)[/tex] C charge where the electric potential is [tex]\(-500\)[/tex] V is [tex]\(96.193\)[/tex] meters.
[tex]\[ V = \frac{k \cdot q}{r} \][/tex]
where:
- [tex]\(V\)[/tex] is the electric potential,
- [tex]\(k\)[/tex] is Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2}\)[/tex]),
- [tex]\(q\)[/tex] is the charge,
- [tex]\(r\)[/tex] is the distance from the charge.
Given:
- [tex]\(V = -500 \, \text{V}\)[/tex],
- [tex]\(q = -5.35 \times 10^{-6} \, \text{C}\)[/tex],
- [tex]\(k = 8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2}\)[/tex].
We need to solve for [tex]\(r\)[/tex]:
[tex]\[ r = \frac{k \cdot q}{V} \][/tex]
Plugging in the given values:
[tex]\[ r = \frac{8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \cdot (-5.35 \times 10^{-6} \, \text{C})}{-500 \, \text{V}} \][/tex]
Simplify the expression:
[tex]\[ r = \frac{(8.99 \times 10^9) \times (-5.35 \times 10^{-6})}{-500} \][/tex]
[tex]\[ r = \frac{-4.8065 \times 10^4}{-500} \][/tex]
[tex]\[ r = 96.193 \, \text{m} \][/tex]
Thus, the distance from the [tex]\(-5.35 \times 10^{-6}\)[/tex] C charge where the electric potential is [tex]\(-500\)[/tex] V is [tex]\(96.193\)[/tex] meters.
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Discover the answers you need at IDNLearn.com. Thanks for visiting, and come back soon for more valuable insights.