Question

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Two point charges, QA = +10C and QB = +12C, are separated by a distance r = 5cm. What is the magnitude of the electric force.

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Answer 1

Answer:

Given :

Two point charges Q₁ = +10C and Q₂ = +12C are separated by a distance of 5 cm (0.05 m).

To Find :

Magnitude of electric force between them.

Solution :

The magnitude of electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them.

This is known as Coulomb's law also known as inverse square law..

Mathematically;

$\dag\:\underline{\boxed{\bf{\gray{F=\dfrac{kQ_1Q_2}{r^2}}}}}$

• r denotes distance

• k denotes coulomb's constant

By substituting the given values;

$\sf:\implies\:F=\dfrac{kQ_1Q_2}{r^2}:$

$\sf:\implies\:F=\dfrac{(9\times10^9)(10)(12)}{(0.05)^2}:$

$\sf:\implies\:F=\dfrac{108\times10^{10}}{25\times10^{-4}}:$

$:\implies\:\underline{\boxed{\bf{\orange{F=4.32\times10^{14}\:N}}}}:$

Hence Proved.