Question

If two charges of 1 coulomb each are placed 1 km apart then the force between them is

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\textsf{ \to There are two charges of 1 Columb .}\\\textsf{ \to They are placed 1km from each other. }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\textsf{ \to The force between them .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

The two charges are 1 C placed 1km apart. According to Columb's Law we know that the force of attraction between two charges is Directly proportional to the product of charges and inversly proportional to the square of distance between them. That is $\sf F \ \propto \ q_1 q_2 \ \& \ F \propto \dfrac{1}{r^2}$ .

$\sf:\implies \pink{ Force \propto \dfrac{ q_1q_2}{r^2} }\qquad \bigg\lgroup \red{\bf From \ above}\bigg\rgroup \\\\\sf:\implies F = k \dfrac{ q_1q_2}{r^2} \\\\\sf:\implies F = \dfrac{1}{4\pi \epsilon_0} \bigg( \dfrac{ q_1q_2}{r^2} \bigg) \\\\\sf:\implies F = \dfrac{ (1C)(1C) }{ (1000m)^2 \times 4\pi \epsilon_0} \\\\\sf:\implies F = 9 \times 10^9 \times \dfrac{1}{ 10^6 } N \\\\\sf:\implies F = 9 \times 10^{(9-6)} N \\\\\sf:\implies \pink{\boxed{\mathfrak{ Force = 9 \times 10^3 Newtons }}}$

$\underline\textsf{\blue{ \therefore Hence the force between them is \textbf{ 9 } \times \textbf{10} ^{\textbf{3}} \textbf{ Newton} .}}$