Question

Two charges of 2 C each are placed 2m apart in free space.The electrostatics force between them will be

Answer 1

Need to FinD :-

• The electrostatic force between charges .

$\red{\frak{Given}}\begin{cases}\sf Two\ equal\ charges \ equal\ to\ 2 C .\\\sf They \ are \ 2m \ apart . \end{cases}$

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 :-

$\sf\dashrightarrow Force = \dfrac{1}{4\pi \epsilon_0}\bigg( \dfrac{q_1q_2}{r^2} \bigg) \\\\\sf\dashrightarrow Force = \dfrac{1}{4\pi\epsilon_0}\bigg( \dfrac{(2C)^2}{(2m)^2}\bigg) \\\\\sf\dashrightarrow Force = \dfrac{4}{4}\times 9 \times 10^9 \\\\\sf\dashrightarrow \underset{\blue{\sf Required\ Answer}}{\underbrace{\boxed{\pink{\frak{ Force = 9\times 10^9 \ N }}}}}$

Hence the force of repulsion ( since charges are same i.e. postive ) is 9 × 10⁹ N .