MCQ
${F_g}$ and ${F_e}$ represents gravitational and electrostatic force respectively between electrons situated at a distance $10\, cm$. The ratio of ${F_g}/{F_e}$ is of the order of
- A${10^{42}}$
- B$10$
- C$1$
- ✓${10^{ - 43}}$
✓
Answer
Correct option: D.
${10^{ - 43}}$
d
(d) Gravitational force between electrons ${F_G} = \frac{{G{{({m_e})}^2}}}{{{r^2}}}$
Electrostatics force between electrons ${F_e} = k.\frac{{{e^2}}}{{{r^2}}}$
$\frac{{{F_G}}}{{{F_e}}} = \frac{{G{{({m_e})}^2}}}{{k.{e^2}}} = \frac{{6.67 \times {{10}^{ - 11}} \times {{(9.1 \times {{10}^{ - 31}})}^2}}}{{9 \times {{10}^9} \times {{(1.6 \times {{10}^{ - 19}})}^2}}}$$ = 2.39 \times {10^{ - 43}}$
(d) Gravitational force between electrons ${F_G} = \frac{{G{{({m_e})}^2}}}{{{r^2}}}$
Electrostatics force between electrons ${F_e} = k.\frac{{{e^2}}}{{{r^2}}}$
$\frac{{{F_G}}}{{{F_e}}} = \frac{{G{{({m_e})}^2}}}{{k.{e^2}}} = \frac{{6.67 \times {{10}^{ - 11}} \times {{(9.1 \times {{10}^{ - 31}})}^2}}}{{9 \times {{10}^9} \times {{(1.6 \times {{10}^{ - 19}})}^2}}}$$ = 2.39 \times {10^{ - 43}}$
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