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Coulomb's Law

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The electrical force, like gravitational force decreases inversely as the square of the distance between charged bodies. This relationship was discovered by Charles Coulomb in the eighteenth century and is called Coulomb's law.

Coulomb's law is a fundamental law of electromagnetism and it seems pertinent to enquire into the degree of confidence with which this law is experimentally known to hold, in particular its inverse square nature. Coulomb's is valid not only over the classical range but deep down in to the quantum domain also. 

Define Coulomb's Law

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It states that, for two charged objects that are much smaller than the distance between them, the force between the two objects varies directly as the product of their charges and inversely as the square of the separation distance. The force acts along a straight line from one charged object to the other.

Coulomb's Law Equation

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Coulomb's law can be expressed as,

F = $\frac{1}{4\pi \epsilon _{0}}$$\frac{q_{1}q_{2}}{r^{2}}$

where,
q1
and q2 are the magnitude of first and second charge
r is the distance between two charges

Coulomb's law Problems

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Solved problems of Coulomb's law are given below :

Solved Examples

Question 1: The distance between electron and proton in an hydrogen atom is $5.3 \times 10^{-11}$m. Obtain the magnitude of the electrical force?

Solution:
 
Given that,
$q_{1}$ = $q_{2}$ = $1.6 \times 10^{-19}$ , r = $5.3 \times 10^{-11}$

F = $\frac{1}{4\pi \epsilon _{0}}$$\frac{q_{1}q_{2}}{r^{2}}$
F = $9 \times 10^{9} \times$ $\frac{(1.6 \times 10^{-19})^{2}}{5.3 \times 10-11^{2}}$
F = $8.2 \times 10^{-8}$ N

 

Question 2: Calculate the force of repulsion if the two charges of magnitude $1.6 \times 10^{-19}$ and $3.2 \times 10^{-19}$ are kept a distance $3.5 \times 10^{-9}$m?
Solution:
 
Given that,
$q_{1}$ = $1.6 \times 10^{-19}$, $q_{2}$ = $3.2 \times 10^{-19}$, r = $3.5 \times 10^{-9}$

F = $\frac{1}{4\pi \epsilon _{0}}$$\frac{q_{1}q_{2}}{r^{2}}$
F = $9 \times 10^{9} \times$ $\frac{1.6×10-19×3.2×10-19}{3.5 \times 10-9^{2}}$
F = $3.76 \times 10^{-11}$ N