Question

Two charges equal to 2 X 10⁻⁶ C are 0.3m apart. If both of them exist inside vacuum, then the force between them is *​

Answer 1

Need to FinD :-

• The force of attraction between charges .

$\red{\frak{Given}}\begin{cases}\sf Two\ equal\ charges \ equal\ to\ 2\times 10^{-6} C .\\\sf They \ are \ 0.3m \ apart . \end{cases}$

We know that if there are two charges of $\sf q_1 \ \& q_2$. And the Distance of seperation between them is r , then the force is equal to ,

$\sf\twoheadrightarrow \boxed{\red{\sf Force = \dfrac{1}{4\pi \epsilon_0}\bigg( \dfrac{q_1q_2}{r^2} \bigg) }}$

Here the value of constant , is 9 × 10⁹ N.m²/C² .

• Substituting the respective values :-

$\sf\dashrightarrow Force = \dfrac{1}{4\pi \epsilon_0}\bigg( \dfrac{q_1q_2}{r^2} \bigg) \\\\\sf\dashrightarrow Force = \dfrac{1}{4\pi\epsilon_0}\bigg( \dfrac{ (2\times 10^{-6})^2}{(0.3m)^2} \bigg)\\\\\sf\dashrightarrow Force = (9\times 10^9) \bigg( \dfrac{4\times 10^{-12} }{0.09 m^2}\bigg) \\\\\sf\dashrightarrow Force = (9\times 10^9) \bigg(\dfrac{4\times 10^{-12+2}}{9} \bigg)\\\\\sf\dashrightarrow Force = (9\times 10^9) \bigg( \dfrac{4\times 10^{-10}}{9}\bigg)\\\\\sf\dashrightarrow Force = 4\times 10^{9-10} N \\\\\sf\dashrightarrow\underset{\blue{\sf Required\ Force}}{\underbrace{\boxed{\pink{\frak{ Force = 0.4 N }}}}}$