Answer

Verified

53.4k+ views

**Hint**So to solve this problem, first of all, we calculate the distance of the charge from the origin, and then the electric field will be calculated by using the formula$E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{q}{{{d^2}}}$. And then the angle will also be calculated as the electric field we will get is negative. And lastly, the vector form of the electric field will be calculated by using the formula$\vec E = E\left[ {\cos \phi \mathop i\limits^ \wedge + \sin \phi \mathop j\limits^ \wedge } \right]$.

Formula used:

Electric field,

$E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{q}{{{d^2}}}$

Here,

$E$, will be an electric field

$q$, will be the charge

$d$, will be the separation between them

${\varepsilon _0}$, permittivity

The vector form of the electric field is given by-

$\vec E = E\left[ {\cos \phi \mathop i\limits^ \wedge + \sin \phi \mathop j\limits^ \wedge } \right]$

**Complete Step by Step Solution**First of all we will calculate the distance of the charge from the origin

$ \Rightarrow d = \sqrt {{{0.30}^2} + {{0.30}^2}} $

And on solving, we get

$ \Rightarrow 0.30 \times \sqrt 2 cm$

Now, since we have distance then we will calculate the electric field.

$E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{q}{{{d^2}}}$

Substituting the values, we get

$ \Rightarrow 9 \times {10^9} \times \left( { - 8 \times {{10}^{ - 9}}} \right) \times \dfrac{1}{{{{0.30}^2} \times 2}}\dfrac{N}{C}$

On solving the above, we get

$ \Rightarrow - 400\dfrac{N}{C}$

Now, we will calculate the electric field strength in the direction of $\theta $

$\tan \theta = \dfrac{{30cm}}{{30cm}}$

And we will get $\tan \theta = 1$

Since it’s making ${45^0}$with the $x - axis$

And from this,

$\cos \theta = \dfrac{1}{{\sqrt 2 }}$ And $\sin \theta = \dfrac{1}{{\sqrt 2 }}$

Since the $E$is negative then the $\theta $will be in the $3rd$quadrant

So from this, we can say

$ \Rightarrow \theta = {180^0} + {45^0}$

In the addition, we get

$ \Rightarrow \theta = {225^0}$

Now, the vector form of the electric field will be

$\vec E = E\left[ {\cos \phi \mathop i\limits^ \wedge + \sin \phi \mathop j\limits^ \wedge } \right]$

Substituting the values, we get

$\vec E = \left( {\dfrac{{400}}{{\sqrt 2 }}} \right)\left[ {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right]\dfrac{N}{C}$

Now we can write it as,

$\vec E = \left( {200\sqrt 2 } \right)\left[ {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right]\dfrac{N}{C}$

**Therefore, the option $\left( c \right)$ is correct.**

**Note**It is the quality of an electric field at a given point or it can likewise be characterized as the power experienced by a unit positive charge set in the electric field. Electric Field intensity is a property of that point in space, whereas F is a property of the point charge placed at the point - a very subtle difference.

Recently Updated Pages

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Which of the following amino acids is an essential class 12 chemistry JEE_Main

Which of the following is least basic A B C D class 12 chemistry JEE_Main

Out of the following hybrid orbitals the one which class 12 chemistry JEE_Main

Other Pages

According to classical free electron theory A There class 11 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main