Short Note

At what separation should two equal charges, 1.0 C each, be placed, so that the force between them equals the weight of a 50 kg person?

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#### Solution

Given:

Magnitude of charges, q_{1} = q_{2} = 1 C

Electrostatic force between them, F = Weight of a 50 kg person = mg = 50 × 9.8 = 490 N

Let the required distance be r.

By Coulomb's Law, electrostatic force,

\[F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}\]

\[ \Rightarrow 490 = \frac{9 \times {10}^9 \times 1 \times 1}{r^2}\]

\[ \Rightarrow r^2 = \frac{9 \times {10}^9}{490}\]

\[\Rightarrow r = \sqrt{\frac{9}{49} \times {10}^8} = \frac{3}{7} \times {10}^4 \text{m} = 4 . 3 \times {10}^3 \text{ m }\]

Concept: Coulomb’s Law

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