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

The electric potential \( V \) at a point due to a point charge \( q \) is given by: \[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r}, \] where: - \( V \) is the electric potential, - \( q = 5 \times 10^{-9} \, \text{C} \) (charge), - \( r \) is the distance of the point from the charge, - \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \). 
Step 1: Rearrange the formula to solve for \( r \). Rearranging: \[ r = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{V}. \] 
Step 2: Substitute the known values. Substitute \( V = 50 \, \text{V} \), \( q = 5 \times 10^{-9} \, \text{C} \), and \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \): \[ r = \frac{9 \times 10^9 \cdot 5 \times 10^{-9}}{50}. \] 
Step 3: Simplify the calculation. \[ r = \frac{45}{50} = 0.9 \, \text{m}. \] Convert \( r \) to centimeters: \[ r = 0.9 \, \text{m} \times 100 = 90 \, \text{cm}. \] 
Final Answer: The distance of \( P \) from the point charge is: \[ \boxed{90 \, \text{cm}}. \]