Question:medium

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, \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} \, {Nm}^2{C}^{-2} \))

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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.
Updated On: Jan 13, 2026
  • 3 cm
  • 9 cm
  • 90 cm
  • 0.9 cm
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The Correct Option is C

Solution and Explanation

The electric potential (\( V_P \)) at a point \( P \) due to a point charge is calculated using the formula: \[V_P = \frac{KQ}{r}\] The constants and variables are defined as: \( K = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, {Nm}^2{C}^{-2} \) (Coulomb's constant) \( Q = 5 \times 10^{-9} \, {C} \) (the point charge) \( r \) (the distance from the charge to point \( P \)) \( V_P = 50 \, {V} \) (the electric potential at point \( P \)) To find the distance \( r \), the formula is rearranged: \[r = \frac{KQ}{V_P}\] Substituting the given values: \[r = \frac{(9 \times 10^9) \times (5 \times 10^{-9})}{50}\] Performing the calculation: \[r = \frac{45 \times 10^0}{50}\] \[r = 0.9 \, {m}\] Therefore, point \( P \) is located 0.9 meters (or 90 cm) from the point charge.
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