Electric potential at a point P due to a point charge 5 × 10⁻⁹ C is 50 V. The distance of P from the charge is: [1/(4πε₀) = 9 × 10⁹ N m² C⁻²]
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Electric potential at a point P due to a point charge 5 × 10⁻⁹ C is 50 V. The distance of P from the charge is: [1/(4πε₀) = 9 × 10⁹ N m² C⁻²]

G1 Physics · Accepted Answer

Electric potential due to a point charge is V = kq/r Therefore, r = kq/V Given: k = 9×10⁹ N m² C⁻² q = 5×10⁻⁹ C V = 50 V So, r = (9×10⁹)(5×10⁻⁹)/50 = 45/50 = 0.9 m = 90 cm Hence option D follows.