Question

Electric potential at a point \( P \) due to a point charge of \( 5 \times 10^{-9} \, \text{C} \) is \( 50 \, \text{V} \). The distance of \( P \) from the point charge is:

• A. 90 cm
• B. 70 cm
• C. 60 cm
• D. 50 cm

Answer 1

The Correct Option is A

To find the distance of point \( P \) from a point charge, we use the formula for electric potential due to a point charge:

\(V = \frac{k \cdot q}{r}\)

where:

• \(V\) is the electric potential at point \( P \).
• \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).
• \(q\) is the charge, \(5 \times 10^{-9} \, \text{C}\).
• \(r\) is the distance we need to find.

Given that the electric potential \( V \) is \(50 \, \text{V}\), we rearrange the formula to solve for \( r \):

\(r = \frac{k \cdot q}{V}\)

Substituting the known values:

\(r = \frac{(8.99 \times 10^9) \cdot (5 \times 10^{-9})}{50}\)

Calculate the numerator:

\(8.99 \times 10^9 \times 5 \times 10^{-9} = 44.95\)

Now, substitute back:

\(r = \frac{44.95}{50}\)

Calculate \( r \):

\(r = 0.899 \, \text{m}\)

Converting meters to centimeters, \( r = 89.9 \, \text{cm} \), which can be approximated to \( 90 \, \text{cm} \).

Thus, the distance of \( P \) from the point charge is 90 cm.