Back to Course

0% Complete
0/0 Steps

• ## Follow us

Lesson 4, Topic 1
In Progress

# Electric Force between Charges

Lesson Progress
0% Complete

A French scientist, Charles Coulomb investigated the nature and magnitude of force between charges and came out with his findings known as Coulomb’s law.

Coulomb’s law states that the force [F] of attraction or repulsion between two charges is directly proportional to the product of their charges Q1 and Q2 and inversely proportional to square of the distance between them.

$$\scriptsize F \propto \normalsize \frac{Q_1 Q_2}{r^2}$$

F = $$\scriptsize k \normalsize \frac{Q_1 Q_2}{r^2}$$

Where k = constant of proportionality

k = $$\frac{1}{4 \pi \varepsilon_0} \scriptsize = 9.0 \: \times \: 10^9 Nm^2 C^{-2}$$

$$\scriptsize \varepsilon_0$$= permittivity of air or free space =  8.85 x 10-12 Fm-1

Materials with high permittivity allow current to pass through them

F = $$\frac{1}{4 \pi \varepsilon_0} \frac{Q_1 Q_2}{r^2}$$

Example

Calculate the force of attraction between two protons of charge 1.6 x 10 -19c each if the distance between them is  5.3 x 10 -11m.

(Take $$\frac{1}{4 \pi \varepsilon_0}$$  = 9 x 109 Nm2 C-2 )

Solution

F = $$\frac{1}{4 \pi \varepsilon_0} \frac{Q_1 Q_2}{r^2}$$

Q1 = Q2 = Q

F = $$\frac{1}{4 \pi \varepsilon_0} \frac{Q . Q}{r^2}$$

= $$\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{r^2}$$

F = $$\scriptsize 9 \times 10^9 \normalsize \frac {\left (1.6 \times 10^{-19} \right)^2}{\left (5.3 \times 10^{-11} \right)^2}$$

= 8.2 x 10 -16 N

error: