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Coulomb's Inverse-Square Law

Introduces magnitude and vector form of Coulomb's law.

When two point charges are brought near each other, they exert an electrostatic force. Charles Augustin de Coulomb discovered that this force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Mathematically, the magnitude of the force $F$ is: $F = k \frac{|q_1 q_2|}{r^2}$

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Formula Card

Coulomb's Law ( Vector Form )

Symbol mapping and units for Coulomb's force.

F_{21} = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r _{21}^{2}} \overset{r}{^}_{21}

Variable Legend

$F_{21}$ : Force on $q_{2}$ due to $q_{1}$ (Newtons, N)
$q_{1}, q_{2}$ : Point charges (Coulombs, C)
$r_{21}$ : Distance between charges (meters, m)
$\overset{r}{^}_{21}$ : Unit vector from $q_{1}$ to $q_{2}$

Constants

$k = \frac{1}{4 π ϵ _{0}} \approx 9 \times 1 0^{9}$ N m $^{2}$ /C $^{2}$
$ϵ_{0} = 8.854 \times 1 0^{- 12}$ C $^{2}$ /(N m $^{2}$ )

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Worked Example

Electric vs Gravitational Force

Compares electrostatic and gravitational forces.

Problem

Coulomb's law for electrostatic force and Newton's law for gravitational force both have inverse-square dependencies on distance.

(a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of the electron and proton due to the electrical force of mutual attraction when they are $1\text{ \AA} \; (10^{-10} \text{ m})$ apart?

Given: $m_p = 1.67 \times 10^{-27}$ kg, $m_e = 9.11 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C.

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Worked Example

Metallic Spheres Contact

Forces after touching charged and uncharged metallic spheres.

Problem

A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B is held by an insulating handle at a distance of 10 cm from A. The repulsion is noted.

Spheres A and B are then touched by identical uncharged spheres C and D respectively. C and D are removed, and B is brought closer to A to a distance of 5.0 cm. What is the expected repulsion of A on the basis of Coulomb's law? Ignore the sizes of A and B.

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Worksheet

Charge Redistribution

Calculate force after spheres touch and separate.

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Let the original charge on sphere A be $q$ and that on B be $q'$ .

At an initial distance $r$ between their centres, the magnitude of the electrostatic force on each is given by $F = \frac{1}{4\pi\varepsilon_0} \frac{qq'}{r^2}$ . When an identical but uncharged sphere C touches A, the charges redistribute symmetrically.

Because the spheres are identical, they share the charge equally, leaving sphere A with a new charge of .

Next, the distance between the centres of spheres A and B is doubled to $2r$ . Substituting the new charge and distance into Coulomb's law, the denominator for the distance squared becomes .

Simplifying the entire expression yields the new expected repulsion force as in terms of the original force $F$ .

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quiz

Coulomb Direction Practice

MCQ testing the unit vector signs.

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Two point charges, $q_1$ and $q_2$ , are both strongly negative. According to the vector formulation of Coulomb's Law, $\vec{F}_{12}$ (the force on $q_1$ due to $q_2$ ) will point in which direction?