Question

A point charge of 0.009µC is placed at the origin. Calculate the intensity of the electric field due to this point charge at point $(\sqrt{2},\sqrt{7},0)$ m.

Answer 1

Hint: Using the position vector and formula for electric field intensity this problem can be solved.

Formula used: The position vector: \[\mathbf{r}=\text{ }x\mathbf{i}+\text{ }y\mathbf{j}+\text{ }z\mathbf{k}~~\] and electric field intensity \[\left( \mathbf{E} \right)\text{ }=\dfrac{KQ}{{{r}^{3}}}\mathbf{r}\].

Complete step by step solution:
The point charge at point $(\sqrt{2},\sqrt{7},0)$ m is given to us. So, substituting the values in position vector we get,
\[\mathbf{r}=\text{ }\sqrt{2}\mathbf{i}+\text{ }\sqrt{7}\mathbf{j}+\text{ 0}\mathbf{k}~~\].
Electric field intensity (E) is given by,
\[\dfrac{KQ}{{{r}^{3}}}\mathbf{r}\]where, r is the position vector, Q is the point charge , r is the magnitude of position vector and \[K=9\times{{10}^{9}}\]
The magnitude of position vector is given by,
\[\begin{align}
  & r=\sqrt{{{\left( \sqrt{2} \right)}^{2}}+{{\left( \sqrt{7} \right)}^{2}}+{{0}^{2}}} \\
 & =\sqrt{2+7+0} \\
 & =\sqrt{9} \\
 & =3\text{ units} \\
\end{align}\]
Now, electric field intensity,
 \[\begin{align}
  & \text{E=}\dfrac{\text{9}\times{{\text{10}}^{9}}\times\text{0}\text{.009}\times{{\text{10}}^{-6}}\left( \sqrt{2}i+\sqrt{7}j+0k \right)}{{{3}^{3}}} \\
 & =\dfrac{81}{27}\left( \sqrt{2}i+\sqrt{7}j+0k \right) \\
 & =3\left( \sqrt{2}i+\sqrt{7}j+0k \right)\text{N/C} \\
\end{align}\]

Therefore, the answer is \[3\left( \sqrt{2}i+\sqrt{7}j+0k \right)\text{N/C}\]

Additional information: Electric field intensity is the force experienced by a unit positive charge placed at that point. It is a vector quantity.

Note: The position vector gives us the point where the point charge is placed at that point for which the force is experienced. So, care must be taken while putting the values in position vector to get the correct magnitude and substitute it in the electric field intensity equation.