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JIPMER 2004
Exam
JIPMER
Year
2004
Subject
Physics
Chapter
Electrostatics

JIPMER 2004 · Physics · Electrostatics

The ratio of electrostatic and gravitational forces acting between...

The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance 5×1011 m5 \times 10^{11} \mathrm{~m}, will be (Charge on electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}, mass of electron =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg}, mass of proton =1.6×10−27 kg=1.6 \times 10^{-27} \mathrm{~kg}, G=6.7×10−11Nm2/kg2)\left.G=6.7 \times 10^{-11} \quad \mathrm{Nm}^2 / \mathrm{kg}^2\right)

Options

  1. A2.36×10392.36 \times 10^{39}
  2. B2.36×10402.36 \times 10^{40}
  3. C2.34×10412.34 \times 10^{41}
  4. D2.34×10422.34 \times 10^{42}

Correct answer

A. 2.36×10392.36 \times 10^{39}

Step-by-step solution

Gravitational force F=GM1M2r2FG=6.7×10−11×9.1×10−31×1.6×10−27(5×10−11)2=3.9×10−47 NF=\frac{G M_1 M_2}{r^2} F_G=\frac{6.7 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-27}}{\left(5 \times 10^{-11}\right)^2} =3.9 \times 10^{-47} \mathrm{~N} Electrostatic force FE=14πε0q1q2r2Fe=9×109×1.6×10−19×1.6×10−19(5×10−11)2=9.22×10−8 N∴ Electrostatic force  Gravitational force =FEFGF_E=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} F_e=\frac{9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(5 \times 10^{-11}\right)^2} =9.22 \times 10^{-8} \mathrm{~N} \therefore \quad \frac{\text{ Electrostatic force }}{\text{ Gravitational force }}=\frac{F_E}{F_G}=9.22×10−83.9×10−47=2.36×1039\begin{aligned} & =\frac{9.22 \times 10^{-8}}{3.9 \times 10^{-47}} \\ & =2.36 \times 10^{39}\end{aligned}$$
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