Question

l Review Part A point charge and a +26.6 μ C point charge equal At what separation is the electrostatic force between a +14.0 in magnitude to 1.60 N ? I1n Submit Memet Provide Feedback Next >

Answer 1

To solve the problem, we will use Coulomb's law:

$F_{e}=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

$k_{e}=8.99x10^{9}N.m^{2}/C^{2}$ Coulomb constant,

$q_{1}$ ($14 \mu C$)and $q_{2}$ ($26.6 \mu C$) are the charges

$F_{e}$ is the force and r is the distance that separates the charges.

Solving the equation for r, we get

$r=\sqrt{k_{e}\frac{q_{1}q_{2}}{F_{e}}}=\sqrt{(8.99x10^{9}N.m^{2}/C^{2})\frac{(14\mu C)(26.6\mu C)}{1.60N}}$

$r=\sqrt{(8.99x10^{9}N.m^{2}/C^{2})\frac{(14x10^{-6} C)(26.6x10^{-6} C)}{1.60N}}$

$r=\sqrt{\frac{3.35N.m^{2}}{1.60N}}=\sqrt{2.094m^{2}}$

$r=1.45m$