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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

BITSAT 2024 · Paper 2 · Physics · Electrostatics

Electric potential at a point 'P' due to a point charge of 5 × 10^...

Electric potential at a point 'P' due to a point charge of 5×10−9C5 \times 10^{-9} \mathrm{C} is 50 V. The distance of ' P ' from the point charge is: (Assume, 14πε0=9×10+9Nm2C−2\frac{1}{4 \pi \varepsilon_0}=9 \times 10^{+9} \mathrm{Nm}^2 \mathrm{C}^{-2} )

Options

  1. A3 cm
  2. B9 cm
  3. C90 cm
  4. D0.9 cm

Correct answer

C. 90 cm

Step-by-step solution

Electric potential at a point P due to a point charge, \left(\because \mathrm{K}=9 \times 10^9\right)V_P=\frac{K Q}{r}\Rightarrow \quad 50=\frac{9 \times 10^9 \times 5 \times 10^{-9}}{\mathrm{r}}\Rightarrow \quad \mathrm{r}=\frac{45}{50}=\frac{9}{10}=0.9 \mathrm{ m}=90 \mathrm{ cm}
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