Question

What is the electric field at a location $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{=}\mathbf{<}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{>}\mathit{m}$, due to a particle with charge +3nC located at the origin?

Short answer

The electric field at a location $\stackrel{\rightharpoonup }{\mathrm{b}}=<-0.1,-0.1,0>m$ is $1.35\times {10}^{3}N/C$ .

Step-by-step solution

Identification of the given data

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 .

Significance of electric field

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.

Determination of the position vector from the origin to the given location <-0.1,-0.1,0>m.

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

Here, $\stackrel{\rightharpoonup }{r}$represents the position vector from the origin to the given location.

Substitute all the values in the above equation.

Determination of the magnitude of the distance from the origin to the given location <-0.1,-0.1,0>m.

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

$\begin{array}{rcl}r& =& \left|\stackrel{\rightharpoonup }{r}\right|\\ & =& \sqrt{{\left(-0.1m\right)}^2+{\left(-0.1m\right)}^2+{\left(0m\right)}^2}\\ & \approx & 0.1414m\end{array}$

Determination of the electric field due to this particle at a location <-0.1,-0.1,0>m .

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 $9\times {10}^{9}N.m^2/C^2$.

Substitute the values in the above equation.

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

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