An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force \(\vec F\) between the two is- (where K=\(\frac{1}{4\pi\epsilon_0}\))
An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force \(\vec F\) between the two is- (where K=\(\frac{1}{4\pi\epsilon_0}\))
\(K\frac{e^2}{r^2}\hat r\)
\(-K \frac{e^2}{r^3}\hat r\)
\(K\frac{e^2}{r^3}\vec r\)
\(-K \frac{e^2}{r^3}\vec r\)
The Correct Option is D
Solution and Explanation
Coulomb force = \(K \frac{e^2}{r^2}(-\hat r)\)
=\(K\frac{e^2}{r^3}\vec r\)
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Concepts Used:
Coulomb’s Law
In 1785, french physicist Charles Augustin de Coulomb coined a tangible relationship in mathematical form between two bodies that have been electrically charged. He represented an equation for the force causing the bodies to attract or repel each other which is commonly known as Coulomb’s law or Coulomb’s inverse-square law.
As per Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges regarded to be point charges.
Coulomb’s Law has an abundant application to modern life, from Xerox machines to laser printers, to powder coating.