Question

Electric potential at a point 'P' due to a point charge of \( 5 \times 10^{-9} \) C is 50 V. The distance of 'P' from the point charge is: (Assume, \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} \, {Nm}^2{C}^{-2} \))

• A. 3 cm
• B. 9 cm
• C. 90 cm
• D. 0.9 cm

Hint:

The electric potential due to a point charge decreases with distance from the charge. The formula \( V = \frac{KQ}{r} \) shows the direct relationship between potential, charge, and distance.

Answer 1

The Correct Option is C

The formula for the electric potential (\( V_P \)) due to a point charge is given by: \[ V_P = \frac{KQ}{r} \] where: \( K = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, {Nm}^2{C}^{-2} \)
\( Q = 5 \times 10^{-9} \, {C} \) (the charge)
\( r \) is the distance from the point charge to the point \( P \)
\( V_P = 50 \, {V} \) (the electric potential at point \( P \))
Rearranging the formula to solve for the distance \( r \):
\[ r = \frac{KQ}{V_P} \] Substituting the known values: \[ r = \frac{(9 \times 10^9) \times (5 \times 10^{-9})}{50} \] \[ r = \frac{45 \times 10^0}{50} \] \[ r = 0.9 \, {m} \] Thus, the distance of point \( P \) from the point charge is 0.9 meters or 90 cm.