Question

An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius. The Coulomb force F between the two is?
(a)- $k\dfrac{{{e}^{2}}}{{{r}^{3}}}\overrightarrow{r}$
(b)- $-k\dfrac{{{e}^{2}}}{{{r}^{3}}}\overrightarrow{r}$
(c)- $k\dfrac{{{e}^{2}}}{{{r}^{2}}}\overrightarrow{r}$
(d)- $-k\dfrac{{{e}^{2}}}{Ar}$

Answer 1

Hint: Here the force is between the charges and the force between two charges is directly proportional to the product of the magnitude of both the charges and the force is inversely proportional to the square of radius between them.

Complete answer: Here the force is between the charges and the force between two charges is directly proportional to the product of the magnitude of both the charges and the force is inversely proportional to the square of radius between them.
In the question, an electron is revolving around the nucleus, this means that one charge is an electron, and the other one is a proton. We know that the charge on the electron is one unit negative so it can be written as -e and the charge on the proton is one unit positive so it can be written as +e, and the radius can be denoted as r. So, we can write the formula as:
$F\propto \dfrac{(-e)(+e)}{{{r}^{2}}}$
Now, removing the proportionality constant we can introduce a constant k. This will be:
$F=-k\dfrac{{{e}^{2}}}{{{r}^{2}}}$
Now, we have define the direction of the forces and the direction will be along the radius that is written as $\overset{\hat{\ }}{\mathop{r}}\,$
This $\overset{\hat{\ }}{\mathop{r}}\,$is equal to:
$\overset{\hat{\ }}{\mathop{r}}\,=\dfrac{\overrightarrow{r}}{r}$
Now, multiplying this direction of force into the value of force, we get:
$F=-k\dfrac{{{e}^{2}}}{{{r}^{2}}}\text{ x }\dfrac{\overrightarrow{r}}{r}$
$F=-k\dfrac{{{e}^{2}}}{{{r}^{3}}}\text{ }\overrightarrow{r}$

Therefore, the correct answer is option (b).

Note: While the force for the substance, then do not forget to mention the direction of force because the force is a vector quantity and in vector quantity, the direction has to be mentioned.