Answer
64.8k+ 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.
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
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)