# The electric charge of an electron is -1.6*10 power-19C. What is the force exerted between two electrons separated by one meter? Please show your calculations.

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### 3 Answers

This is a simple application of Coulomb's Law. It's directly analogous to the Universal Gravitation equation, because both follow the Inverse Square law, so one should look familiar if you've seen the other.

The law states:

`F = (kqQ)/r^2 `

Where q and Q are two electrical charges, r is the distance between them, and k is the Coulomb constant. From here it's just a matter of plugging in the values.

K has a value of 8.987 x 10e9 (or, if you're doing things in a hurry, the simple "9e9" is a good way to remember it)

q and Q, in this case, are both the charge of one electron, 1.602 x 10e-19.

r is 1 meter, so it's not relevant to the math. Thus our force will be kqQ.

(8.987 x 10e9) (1.602 x 10e-19) (1.602 x 10e-19)

23.06 x 10e-29

Adjust for scientific notation and you get **2.306 x 10e-28N**

According to Coulomb's law. the electric force between two charged particles is directly proportional to the product of the charges and inversely proportional to the distance between them squared. The mathematical expression of the law is as follows:

F = K Q1Q2/r^2

Here K = 1/4πԐ0, is a constant and its value depends on the medium surrounding the electric charges. If the charges are in a vacuum, then K = 9 × 10^9 Nm^2/C^2

For two electrons separated by a distance of 1 m the calculation would be:

F = (9 × 10^9 Nm^2/C^2)(1.6 × 10^-19 C^2)^2/(1 m)^2

F = 14.4 × 10^-10 N = 1.44 × 10^-9 N

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1.6 needs to be squared as well, before multiplying by 9.

Electric force acting on a point charge q1 in presence of a point charge q2 is given by Coulomb's law:

F = k*(q1)*(q2) / r²

where r is the distance between the point charges and k = Coulomb's constant = 9 * 10⁹ N·m²/C²

Here, r = 1 m

So, the required force =

(9 × 10⁹ N·m²/C²)(1.6 × 10ˉ¹⁹ C)(1.6 × 10ˉ¹⁹ C) / 1² m²

= **2.3 × 10ˉ²⁸ N**