Question

a two charges q1 and q2 of their magnitude 2×10^-6 and 4×10^-6 are separated by a finite distance 12cm.then find the magnitude of force of attraction acting between them​

Answer 1

ANSWER:

• Magnitude of force of attraction acting between them​ = 5 N.

GIVEN:

• Two charges $\sf q_1$ and $\sf q_2$  of their magnitude $\sf 2\times10^{-6}\ C$ and $\sf4\times10^{-6}\ C$ are separated by a finite distance 12 cm.

TO FIND:

• Magnitude of force of attraction acting between them​.

EXPLANATION:

$\boxed{\bold{\large{\gray{F = \dfrac{1}{4\pi \epsilon_o}\dfrac{q_1\ q_2}{r^2}}}}}$

$\sf q_1 = 2\times10^{-6}\ C$

$\sf q_2=4\times10^{-6}\ C$

$\sf r=12\times10^{-2}\ m$

$\sf F = 9\times10^{9}\times\dfrac{2\times10^{-6}\times4\times10^{-6}}{(12\times10^{-2})^2}$

$\sf F = 9\times10^{9}\times\dfrac{8\times10^{-12}}{144\times10^{-4}}$

$\sf F = \dfrac{72\times10^{-3}}{144\times10^{-4}}$

$\sf F = \dfrac{1\times10^{1}}{2}$

$\sf F = 0.5\times10$

F =  5 N

HENCE THE ATTRACTIVE FORCE = 5 N

COULOMB's LAW:

• Coulomb's law of electrostatics states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

• Coulomb's law is similar to the Newton's law of gravitation.

$\boxed{\bold{\large{\gray{F \propto q_1\ q_2}}}}}$

$\boxed{\bold{\large{\gray{F \propto\dfrac{1}{r^2}}}}}$

$\boxed{\bold{\large{\gray{F \propto\dfrac{q_1\ q_2}{r^2}}}}}$

$\boxed{\bold{\large{\gray{F = \dfrac{1}{4\pi \epsilon_o}\dfrac{q_1\ q_2}{r^2}}}}}$

$\sf Here \ \epsilon_o\ is\ permittivity\ of\ free\ space.$