Answer :
Answer:
[tex]\mathbf{1.51\times10^{-15}N}[/tex]
Explanation:
The computation of the weight of the paper in newtons is shown below:
On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.
Therefore the paper charge is
[tex]q_{paper}=-4.1\times10^{-15}C[/tex]
Now the distance from the charge is
[tex]r=1cm=0.01m[/tex]
Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.
[tex]mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}[/tex]
[tex]\Rightarrow W=mg[/tex]
[tex]=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}[/tex]
[tex]=\mathbf{1.51\times10^{-15}N}[/tex]