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

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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 , 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:

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat

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