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Question
Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges, so that the force between them equals the weight of a 50 kg person?
Solution
Let the magnitude of each charge be q.
Separation between them, r = 1 m
Force between them, F = 50 × 9.8 = 490 N
By Coulomb's Law force, \[F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}\]
\[\Rightarrow 490 = 9 \times {10}^9 \times \frac{q^2}{1^2}\]
\[ \Rightarrow q^2 = 54 . 4 \times {10}^{- 9} \]
\[ \Rightarrow q = \sqrt{54 . 4 \times {10}^{- 9}} = 23 . 323 \times {10}^{- 5} C\]
\[\text{ Or q }= 2 . 3 \times {10}^{- 4} C\]
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F = `(Qq)/r^2 hatr`
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1 esu of charge = 1 (dyne)1/2 cm.
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