Question

A point charge of $100\mu C$ is placed at $3\widehat{i}+4\widehat{j}\text{ }m$. Find the electric field intensity due to this charge at a point located at $9\widehat{i}+12\widehat{j}\text{ }m$:
$\begin{aligned} & \text{A}\text{. }800V{{m}^{-1}} \\\ & \text{B}\text{. }9000V{{m}^{-1}} \\\ & \text{C}\text{. }2250V{{m}^{-1}} \\\ & \text{D}\text{. }4500V{{m}^{-1}} \\\ \end{aligned}$

Answer 1

Electric field is an electric property associated with each and every point in space when some charge is present in any of its forms. Electric field intensity at a point can be calculated by finding the amount of force experienced by a unit positive charge placed at that point. For calculating the electric field intensity at the given point, we need to first calculate the distance between the two charges and then we can apply the formula of electric field intensity.
Formula used:
Electric field intensity due to a point charge,
$E=\dfrac{kq}{{{r}^{2}}}$

Complete step by step answer:
The 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 the electric field is an electric property associated with each and every point in space when some charge is present in any of its forms. The magnitude and the direction of the electric field vector are expressed by the value, called 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 of space around a charge in which it exerts an electrostatic force on other charges while the strength of the electric field at any point in space is called as electric field intensity. The 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 a point charge of $100\mu C$ being placed at $3\widehat{i}+4\widehat{j}\text{ }m$ and we have to calculate the electric field intensity due to this charge at a point located at $9\widehat{i}+12\widehat{j}\text{ }m$.
Radius vector is given as,
$\left( {{x}_{2}}-{{x}_{1}} \right)\widehat{i}+\left( {{y}_{2}}-{{y}_{1}} \right)\widehat{j}$
We have,
$\begin{aligned} & {{x}_{1}}=3 \\\ & {{x}_{2}}=9 \\\ & {{y}_{1}}=4 \\\ & {{y}_{2}}=12 \\\ \end{aligned}$
Thus,
$\begin{aligned} & \overrightarrow{r}=\left( 9-3 \right)\widehat{i}+\left( 12-4 \right)\widehat{j} \\\ & \overrightarrow{r}=6\widehat{i}+8\widehat{j} \\\ \end{aligned}$
Magnitude of radius vector,
$\left| \overrightarrow{r} \right|=\sqrt{{{6}^{2}}+{{8}^{2}}}=\sqrt{100}=10m$
Now,
Electric field intensity due to a point charge is given as,
$E=\dfrac{kq}{{{r}^{2}}}$
Where,
$k$ is the coulomb’s law constant
$q$ is the value of charge
$r$ is the distance between point and the charge
Putting the values,
$\begin{aligned} & k=9\times {{10}^{9}} \\\ & q=100\times {{10}^{-6}} \\\ & r=10 \\\ \end{aligned}$
We get,
$\begin{aligned} & E=\dfrac{9\times {{10}^{9}}\times 100\times {{10}^{-6}}}{{{10}^{2}}}=\dfrac{9\times {{10}^{5}}}{100}=9\times {{10}^{3}} \\\ & E=9000V{{m}^{-1}} \\\ \end{aligned}$
The required the electric field intensity is $9000V{{m}^{-1}}$

Hence, the correct option is B.

Note: In the formula for electric field intensity, we use the distance between the two charges. The position of the charges does not matter while calculating the electric field intensity. But when we have to calculate the electric field vector, the positions of the respective charges play a significant role in determining the value and direction of the net electric field at a particular point.