91Ó°ÊÓ

The given data can be listed below as:

• The location of a charged particle is .
• The electric charge on a particle is $q=+3nC$.
• The location of another point is .

The electric field at a specific point can be obtained with the help of Coulomb’s law. If the magnitude of the electric charge is more, it means the intensity of the electric field would be more.

The expression of the position vector from the origin to the given location can be expressed as:

Here, $\stackrel{â‡\P Ä}{r}$represents the position vector from the origin to the given location.

Substitute all the values in the above equation.

The magnitude of the distance from the origin to the given location can be calculated as:

$\begin{array}{rcl}r& =& \left|\stackrel{â‡\P Ä}{r}\right|\\ & =& \sqrt{{\left(-0.1m\right)}^2+{\left(-0.1m\right)}^2+{\left(0m\right)}^2}\\ & ≈& 0.1414m\end{array}$

The expression of the electric field due to this particle at a location can be expressed as:

$E=k\left(\frac{q}{r^2}\right)$

Here, E represents the required electric field due to this particle at a locationand k represents Coulomb’s constant, whose value is $9×{10}^{9}N.m^2/C^2$.

Substitute the values in the above equation.

$\begin{array}{rcl}E& =& \left(9×{10}^{9}N.m^2/C^2\right)\left[\frac{3nC×\left({\displaystyle \frac{{10}^{-9}C}{1nC}}\right)}{{\left(0.1414m\right)}^2}\right]\\ & =& 1.35×{10}^{3}N/C\end{array}$

Hence, the electric field due to this particle at a location is $\begin{array}{rcl}& & 1.35×{10}^{3}N/C\end{array}$.