Question

An infinite number of charges each of magnitude q are placed on x-axis at distances of 1, 2, 4, 8, ….… meter from the origin. The intensity of the electric field at origin is
(A) ${\displaystyle \frac{q}{3\pi {\epsilon }_0}}$
(B) ${\displaystyle \frac{q}{6\pi {\epsilon }_0}}$
(C) ${\displaystyle \frac{q}{2\pi {\epsilon }_0}}$
(D) ${\displaystyle \frac{q}{4\pi {\epsilon }_0}}$

Answer 1

Hint:In this problem,we are going to apply the concept of electric field at a distance r due to a point charge q. The electric field intensity at a point r from a charge q is given by the relation$$E={\displaystyle \frac{1}{4\pi {\in }_0}}\cdot {\displaystyle \frac{q}{r^2}}$$.Find the sum of the intensities by all the charges of magnitude q.

Complete step by step answer:
Let there be a source charge placed at the origin. We know that the electric field intensity by a point charge placed at a distance of r from the source charge is given by the relation
$$E={\displaystyle \frac{1}{4\pi {\in }_0}}\cdot {\displaystyle \frac{q}{r^2}}$$ or
$$E={\displaystyle \frac{kq}{r^2}}$$
Where, $$k={\displaystyle \frac{1}{4\pi {\in }_0}}$$
As it is given that an infinite number of charges each of magnitude q are placed on the x-axis at distances of 1, 2, 4, 8, ….… meter from the origin. Therefore, the electric field at origin is equal to the algebraic sum of the individual charges. So,
$E_{origin}={\displaystyle \frac{kq}{1^2}}+{\displaystyle \frac{kq}{2^2}}+{\displaystyle \frac{kq}{4^2}}+{\displaystyle \frac{kq}{8^2}}+...................$
$\Rightarrow E_{origin}=kq\left({\displaystyle \frac{1}{1^2}}+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{4^2}}+{\displaystyle \frac{1}{8^2}}+...................\right)$
$\Rightarrow E_{origin}=kq\left({\displaystyle \frac{1}{1^2}}+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{2^4}}+{\displaystyle \frac{1}{2^6}}+...................\right)$
The series $\left({\displaystyle \frac{1}{1^2}}+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{2^4}}+{\displaystyle \frac{1}{2^6}}+...................\right)$ is a geometric progression (GP) containing infinite series whose sum is given by ${\displaystyle \frac{a}{1-r}}$
Here, a = 1 and r = ${\displaystyle \frac{1}{2^2}}={\displaystyle \frac{1}{4}}$
${\displaystyle \frac{1}{1^2}}+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{2^4}}+{\displaystyle \frac{1}{2^6}}+...................={\displaystyle \frac{1}{1-{\displaystyle \frac{1}{4}}}}={\displaystyle \frac{4}{3}}$
$$\therefore E_{origin}=kq{\displaystyle \frac{4}{3}}={\displaystyle \frac{q}{3\pi {\in }_0}}$$

Hence, the correct option is (A).

Note: The electric field intensity at any point is the strength of the electric field at that point. Electric field intensity is defined as the force experienced by unit positive charge placed at that point. The electric field $$\overrightarrow{E}$$ is created by a source charge placed in the vicinity of the given charge. This source charge could be a single charge or a group of discrete charges or any continuous distribution of charge. The electric field intensity at any point due to a group of point charges is equal to the vector sum of the electric field intensities due to the individual charges at the same point.