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

Answer:

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}F

e

= W, Weight.

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

F_{e}F

e

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

r

2

q

1

q

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.

Answer 2

At 3585.6 m  distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight.

Explanation:

Step 1:

Given data :

                   q₁ = q₂ = 1 C.

We know that,

                      $\frac{1}{4 \pi \varepsilon_{0}}=9.0 \times 10^{9} \mathrm{N}-\mathrm{m}^{2} / \mathrm{C}^{2}$

The electrostatic force should be known here = W, Weight.

Step 2:

Now, two consecutive charges between the force of attraction,  

                      $\mathrm{F}_{\mathrm{e}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$    

Let's presume body weight i.e. = 1000 N  

                     $700=\left(9 \times 10^{9}\right) \frac{1 \times 1}{r^{2}}$

                     $700=\left(9 \times 10^{9}\right) \frac{1}{r^{2}}$

             $700 \times r^{2}=\left(9 \times 10^{9}\right)$

                      $r^{2}=\frac{\left(9 \times 10^{9}\right)}{700}$

                      $r^{2}=0.012857 \times 10^{9}$

                       $r=\sqrt{0.012857 \times 10^{9}}$

                       r = 3585.6 m

The distance at which the two 1 C charges must be held is 3585.6 m. So that, my weight is equal to the force between them.