At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight ?
Solution
Given : \[q_1 = q_2 = 1 C\]
By Coulomb's law, the force of attraction between the two charges is given by
\[F = \frac{1}{4\pi \in_0}\frac{q_1 q_2}{r^2}\]
\[ = \frac{9 \times {10}^9 \times 1 \times 1}{r^2}\]
However, the force of attraction is equal to the weight (F = mg).
\[\therefore \text{ mg } = \frac{9 \times {10}^9}{r^2}\]
\[ \Rightarrow r^2 = \frac{9 \times {10}^9}{\text{ m } \times 10} = \frac{9 \times {10}^8}{\text{ m } } (\text{ Taking } g = 10 \text{ m } / s^2 )\]
\[ \Rightarrow r^2 = \frac{9 \times {10}^8}{\text{ m } }\]
\[ \Rightarrow r = \frac{3 \times {10}^4}{\sqrt{\text{ m } }}\]
Assuming that m = 81 kg, we have :
\[r = \frac{3 \times {10}^4}{\sqrt{81}}\]
\[ = \frac{3}{9} \times {10}^4 \ \text{ m }\]
\[ = 3333 . 3 \ \text{ m } \]
∴ The distance r is 3333.3 m.
