Question

A particle of charges ${q_0}$​ is moved around a charge +q along the semicircle path of radius r from A to B (see figure). The work done by the Coulomb force is:

A) $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_0}q}}{r}$
B) $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2{q_0}q}}{r}$
C) $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2{q_0}q}}{r}.\pi r$
D) $Zero$

Answer 1

The Coulomb force is also called the electrostatic force and is defined as the force of the attraction or repulsion of the particles because of the electric charge.

Complete step by step solution:
Given data:
A particle of charges ${q_0}$ is moved around a charge +q along the semicircle path of radius r.
Then work done by the Coulomb force =?
We know that the work done by the Coulomb force is given by the formula, $w = qV$
The charge ${q_0}$ is moved and the potential due to the charge +q at a distance of r is V.
Thus $V = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{r}$
Thus, Potential at A and B is the same.
Thus substituting the values of q and V, we get,
$W = {q_0} \Delta V$
$\Rightarrow W = 0$

Hence the correct option is D.

Note: 1. Coulomb’s law states that the force of the attraction or the repulsion between the charged objects will be directly proportional to the product of the charges and is inversely proportional to the square of the distance between the objects.
2. The property of matter that causes it to experience the force when placed in the electromagnetic field is called the Electric Charge. It is of two types namely positive charge and negative charge. The like charges will repel each other and the unlike charges will attract each other.
3. When the matter consists of more number of protons than electrons, then we can say that the electric charge is the positive charge. Similarly, when the matter consists of more electrons than protons, then we can say that the electric charge is the positive charge.