If two such generic humans each carried 2.0 C coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650 N weight?

Answer :

Given :

Charge on human 1 , [tex]q_1=2\ C[/tex] .

Charge on human 2 , [tex]q_2=-2\ C[/tex] .

Force of attraction between them , F = 650 N .

To Find :

Force between them , F =- 650 N .( force between opposite charge is -ve )

Solution :

We know , electrostatics force between two charge is given by :

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

Here k is universal gravitational constant , [tex]k=6.67408 \times 10^{-11}\ m^3\ kg^{-1}\ s^{-2}[/tex] .

Putting all given value in above equation we get :

[tex]-650=\dfrac{6.67408 \times 10^{-11}\times 2\times (-2)}{r^2}\\\\r=\sqrt{\dfrac{6.67\times 10^{-11}\times 2\times 2}{650}}\\\\r=6.40\times 10^{-7}\ m[/tex]

Hence , this is the required solution .

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