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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Electrostatics

JIPMER 2018 · Physics · Electrostatics

If point charges Q_1=2 × 10^ -7 C and Q_2=3 × 10^ -7 C are at 30 cm...

If point charges Q1=2×10−7C\mathrm{Q}_1=2 \times 10^{-7} \mathrm{C} and Q2=3×10−7C\mathrm{Q}_2=3 \times 10^{-7} \mathrm{C} are at 30 cm separation, then find electrostatic force between them

Options

  1. A2×10−3 N2 \times 10^{-3} \mathrm{~N}
  2. B6×10−3 N6 \times 10^{-3} \mathrm{~N}
  3. C5×10−3 N5 \times 10^{-3} \mathrm{~N}
  4. D1×10−3 N1 \times 10^{-3} \mathrm{~N}

Correct answer

B. 6×10−3 N6 \times 10^{-3} \mathrm{~N}

Step-by-step solution

∵\because Electrostatic force, F=kQ1Q2r2\mathrm{F}=\mathrm{k} \frac{\mathrm{Q}_1 \mathrm{Q}_2}{\mathrm{r}^2} =9×109×2×10−7×3×10−7(30×10−2)2=6×10−3 N=9 \times 10^9 \times \frac{2 \times 10^{-7} \times 3 \times 10^{-7}}{\left(30 \times 10^{-2}\right)^2}=6 \times 10^{-3} \mathrm{~N}
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