At What Distance Should Two Charges, Each Equal to 1 C, Be Placed So that the Force Between Them Equals Your Weight ? - Physics

Short Note

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.

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 4 The Forces
Exercise | Q 3 | Page 63