Question:

The value of electric potential at a distance of 9 cm from the point charge \(4 \times 10^{-7} \, \text{C}\) is
\[ \text{Given} \, \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \]

Updated On: Dec 9, 2024
  • \(4 \times 10^2 \, \text{V}\)
  • \(44.2 \, \text{V}\)
  • \(4.4 \times 10^5 \, \text{V}\)
  • \(4 \times 10^5 \, \text{V}\)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

Electric potential (V) due to a point charge (q) at a distance (r) is given by:

$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$

Given: $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N m}^2 \, \text{C}^{-2}$ q = 4 × 10-7 C r = 9 cm = 0.09 m

$V = (9 \times 10^9 \, \text{N m}^2 \, \text{C}^{-2}) \frac{(4 \times 10^{-7} \, \text{C})}{0.09 \, \text{m}} = 4 \times 10^4 \, \text{V}$

Was this answer helpful?
0
0

Top Questions on Thermodynamics

View More Questions