Question

Two point charges $-q$ and $\dfrac{+q}{2}$ are situated at the origin and the point $(a,0,0)$ respectively. The point along the axis, where the electric field vanishes is
$\text{A}\text{. }x=\sqrt{2}a$
$\text{B}\text{. }x=\dfrac{a}{\sqrt{2}}$
$\text{C}\text{. }x=\dfrac{\sqrt{2}a}{\sqrt{2}-1}$
$\text{D}\text{. }x=\dfrac{\sqrt{2}a}{\sqrt{2}+1}$

Answer 1

Hint: The point where the electric field due to both charges vanishes can be found out by calculating the strength of the electric field at a random point along the axis where the charges are situated. Equating the value of the electric field due to each charge, we will get the desired point.
Formula used:
$E=\dfrac{kQ}{{{r}^{2}}}$

Complete step by step answer:
Electric field is described as the region around a charged particle or object within which another charged particle or object will experience some force. We can say that an electric field is an electric property associated with each and every point in space when some charge is present in any of its forms. Magnitude and the direction of electric field vector are expressed by the value, called as electric field intensity, or electric field strength or simply the electric field. The basic difference between electric field, or electric field vector and electric field intensity is that the electric field is a specific region or space around a charge in which it exerts an electrostatic force on other charges while the strength of electric field at any point in space is called as electric field intensity. Electric field is a vector quantity.
The magnitude of electric field $E$ produced by a point charge with a charge of magnitude$Q$, at a distance$r$away from the point charge, is given by:
$E=\dfrac{kQ}{{{r}^{2}}}$
Where,
$k$ is a constant having value of $8.99\times {{10}^{9}}\dfrac{N{{m}^{2}}}{{{C}^{2}}}$
We are given that two point charges $-q$ and $\dfrac{+q}{2}$ are situated at the origin and the point $(a,0,0)$respectively.

Let’s assume that the point P where the electric field due to both charges vanishes is situated at a distance $x$ from the origin along the x-axis.
Electric field at point $x$ due to charge $-q$ situated at origin,
${{E}_{1}}=\dfrac{-kq}{{{(a+x)}^{2}}}$
Electric field at point$x$due to charge$\dfrac{+q}{2}$situated at$(a,0,0)$,
${{E}_{2}}=\dfrac{k\dfrac{q}{2}}{{{x}^{2}}}=\dfrac{kq}{2{{x}^{2}}}$
Direction of ${{E}_{1}}$ will be towards negative x-axis, because of attractive force, and direction of ${{E}_{2}}$will be towards positive x-axis, because of repulsion.
For electric field to be zero at point P,
${{E}_{1}}+{{E}_{2}}=0$
We get, $\dfrac{kq}{{{(a+x)}^{2}}}=\dfrac{kq}{2{{x}^{2}}}$
$\begin{aligned} & 2{{x}^{2}}={{(a+x)}^{2}} \\\ & \sqrt{2}x=a+x \\\ & x=\dfrac{a}{\sqrt{2}-1} \\\ \end{aligned}$
Since, $x$ is calculated from point $(a,0,0)$, value of $x$ from origin will be,
$\dfrac{a}{\sqrt{2}-1}+a=\dfrac{\sqrt{2}a}{\sqrt{2}-1}$
The point along the axis where the electric field will vanish is situated at a distance $\dfrac{\sqrt{2}a}{\sqrt{2}-1}$ from the origin along the positive x-axis.
Hence, the correct option is C.
Note: While calculating the strength of the electric field at point P, keep in mind the direction of the electric field vector. To get the value of electric field zero at point P, the sum of electric fields due to individual charge would be zero. The distance should be calculated taking the reference point as the origin.