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SS3: PHYSICS - 1ST TERM

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Lesson 4, Topic 1
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Electric Force between Charges

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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.

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

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

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