Question

Define and give Coulomb's law.

Answer 1

ANSWER


COULOMBS LAW

The force of attraction and repulsion between charges
is directly proportional to the magnitude of two charges and inversely proportional to square of distance between them

f=Q1Q2/R^2


where

Q ------charge

R-------distance

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Answer 2

EXPLANATION.

Electric force (Electrostatics force) between two stationary point charges.

There are two forces Q₁  and  Q₂ separate by distance r.

⇒ F ∝ q₁q₂.

⇒ F ∝ 1/r² (inverse square law).

⇒ F = Kq₁q₂/r².

⇒ k = proportionality constant.

⇒ k = Electrostatic force constant.

⇒ k = 9.0 x 10⁹ nm²/c²  for air/free space / vacuum.

Coulomb's law in vector form.

$\sf \implies \vec{F_1_2} = \dfrac{kq_1q_2}{r^{2} } (\hat{r_1 - \vec{r_2}})$

$\sf \implies \hat{A} = \dfrac{\vec{A}}{|A|}$

$\sf \implies |r| = | \vec{r_1} - \vec{r_2} |$

$\sf \implies \vec{F_1_2} = \dfrac{kq_1q_2}{|r|^{2} } \times \dfrac{(\vec{r_1} - \vec{r_2})}{|\vec{r_1} - \vec{r_2}|}$

$\sf \implies \boxed{\vec{F_1_2} = \dfrac{kq_1q_2}{|\vec{r_1} - \vec{r_2}|^{3} } (\vec{r_1} - \vec{r_2})}$

$\sf \implies \vec{F_2_1} = \dfrac{kq_1q_2}{r^{2} } (\hat{\vec{r_2} - \vec{r_1})}$

$\sf \implies \vec{F_2_1} = \dfrac{kq_1q_2}{|r^{2} |} \times \dfrac{\vec{r_2} - \vec{r_1}}{|\vec{r_2} - \vec{r_1}|}$

$\sf \implies \boxed{\vec{F_2_1} = \dfrac{kq_1q_2}{|\vec{r_2} - \vec{r_1}|^{3} } (\vec{r_2} - \vec{r_1})}$