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(|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)(Nm^(2))/(C^(2)) (|-8.00x10^(-5)C*-8.00x10^(-5)C|)/((0.20m)^(2) )`

`F = 9.00x10^(9)(Nm^(2))/(C^(2))(1.6x10^-7(C^(2) )/(m^(2) ) })\\F = 1440N`