Question

Two identical charges, each -8.00 E-5 C, are separated by a distance of 20.0 cm (100 cm = 1 m). What is the force of repulsion? Coulomb's constant is 9.00 E9 N*m2/C2
Remember to identify all data (givens and unknowns), list equations used, show all your work, and include units and the proper number of significant digits to receive full credit.

Answer 1

F = 1440 N. The repulsion force between two identical charges, each -8.00x10⁻⁵C separated by a distance of 20.0 cm is 1440 N.

The easiest way to solve this problem is using Coulomb's Law given by the equation $F=k\frac{|q_{1}*q_{2}|}{r^{2} }$, where k is the constant of proportionality or Coulomb's constant, q₁ and q₂ are the charges magnitude, and r is the distance between them.

We have to identical charges of -8.00x10⁻⁵C, are separated by a distance of 20.0 cm, and we need to know the force of repulsion between the charges.

First, we have to convert 20.0 cm to meters.

(20.0 cm x 1m)/100cm = 0.20 m

Using the Coulomb's Law equation:

$F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}} \frac{|-8.00x10^{-5}C*-8.00x10^{-5}C|}{(0.20m)^{2} }$

$F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}}(1.6x10^-7\frac{C^{2} }{m^{2} } })\\F = 1440N$