Question

Two positively charged particles, each of mass 1.7 x 10⁻²⁷ kg and carrying a charge of 1.6 x10⁻¹⁹ C are placed at a distance d apart. If each experiences a repulsive force equal to its weight, find the value of d.

Answer 1

Step 1: Given.

$m=1.7x{10}^{-27}kg$

$q=1.6x{10}^{-19}C$

Step 2: Find out.

We have to determine the value of d.

$F_{electrostatic}=F_{gravity}$

$\frac{K{q}^2}{d^2}=mg$

or $\frac{9\times {10}^{19}{(1.6\times {10}^{-19})}^2}{d^2}=1.7\times {10}^{-27}\times 10$

$⟹d=0.11m=11cm$