Question

Two equal charges are placed at a separation of 1 M what should be the magnitude of the charge so that the force between them equals the weight of a 50 kg person

Answer 1

weight of a 50kg person=50×9.8=490N
by coloumb's law
F=kq1q2/r²
here F=490,q1=q2=q and r=1
490N=q²(9×109 N·m2·C−2),where k=9×109 N·m2·C−2

now....
q²=490/9×109
q=2.34×10^(-8)

Answer 2

Answer:

The charge is $q=2.3\times10^{-4}\ C$.

Explanation:

Given that,

Weight W=$mg= 50\times9.8$

Distance d = 1 m

Charge $q_{1}=q_{2}=q$

The force between the charges is

$F= \dfrac{kq_{1}q_{2}}{d^2}$.....(I)

Here, F = force

q = charge

d = distance

Put the value in equation (I)

$50\times9.8=\dfrac{9\times10^{9}q^2}{1}$

$q^2=\dfrac{50\times9.8}{9\times10^{9}}$

$q=\sqrt{\dfrac{50\times9.8}{9\times10^{9}}}$

$q=2.3\times10^{-4}\ C$

Hence, The charge is $q=2.3\times10^{-4}\ C$.