Question

At what distance should two charges, each equal to 1 C, be placed so that the force between them equals to your weight?
Concept of Physics - 1 , HC VERMA , Chapter "The Force".

Answer 1

Hello Dear.

Given quantities are :- 
q₁ = q₂ = 1 C.

we know the value of 1/ 4πε₀ = 9.0 x 10⁹ N-m²/ C²

Here we should know that Electrostatic force $F_{e}$ = W, Weight.

Now, the force of attraction b/w two consecutive charges,

$F_{e}$ = (1/ 4πε₀ ). $\frac{q _{1} q_{2} }{r^{2} }$ 

Let assume the weight of the body i.e me = 1000 N 
700 = 9 x 10⁹ . (1 x 1 )/r²
r² = 9x 10⁹ / 1000
r² = 9 x10⁶
r = √ (9 x10⁶ )

r = 3000 m . 


3000 m. is the distance at which the two charges of 1 C must be kept. so that the force b/w them is equal to my weight.


Hope it Helps. :-)

Answer 2

Explanation:

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