Question

At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight ?

Answer 1

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.