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# Electric field at a point $\left( {30,30,0} \right)$ due to a point charge of $8 \times {10^{ - 3}}\mu c$ placed at origin will be (coordinates are in cm)$\left( a \right){\text{ 8000N/C}}$$\left( b \right){\text{ 4000}}\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){\text{N/C}}$$\left( c \right){\text{ 200}}\sqrt 2 \left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){\text{N/C}}$$\left( d \right){\text{ 400}}\sqrt 2 \left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){\text{N/C}}$

Last updated date: 13th Jun 2024
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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.