Q68P - Fundamentals Of Physics

Question

In Fig. 22-65, eight particles form a square in which distance$\mathit{d}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{2}\mathbf{.0}\mathbf{ }\mathbf{\text{cm}}$. The charges are,${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{3}\mathit{e}$, ${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{+}\mathit{e}$, ${\mathit{q}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{5}\mathit{e}$,${\mathit{q}}_{\mathbf{4}}\mathbf{=}\mathbf{-}\mathbf{2}\mathit{e}$ , ${\mathit{q}}_{\mathbf{5}}\mathbf{=}\mathbf{+}\mathbf{3}\mathit{e}$, ${\mathit{q}}_{\mathbf{6}}\mathbf{=}\mathbf{+}\mathit{e}$,${\mathit{q}}_{\mathbf{7}}\mathbf{=}\mathbf{-}\mathbf{5}\mathit{e}$ and ${\mathit{q}}_{\mathbf{8}}\mathbf{=}\mathbf{+}\mathit{e}$. In unit-vector notation, what is the net electric field at the square’s center?

Step-by-Step Solution

Answer

The net electric field at the square’s centre is.$(1.08\times {10}^{-5}\text{N/C}) \widehat{i}$

Step 1: The given data

 

1. Eight charged particles form a square in which distance, $d=2.0\text{cm}$(as shown in fig.).
2. The values of the charges,$q_1=+3e$,,$q_2=+e$,$q_1=+3e$,$q_3=-5e$,$q_4=-2e$,$q_5=+3e$$q_6=+e$,$q_7=-7e$and$q_8=+e$.

Step 2: Understanding the concept of the electric field

Using the basic concept of the electric field on a point due to a particle at other points, we can get the individual electric fields of the charges. Then, adding them up will give us the net electric field of the charges on the point.

 

Formula:

The electric field at a point to a charge, $\overrightarrow{E}=\frac{q}{4\pi {\epsilon }_o{r}^2}\widehat{r}$                                        (i)

Where, r = The distance of field point from the charge

            q = charge of the particle

 

Step 3: Calculation of the net electric field

Most of the individual fields, caused by diametrically opposite charges, will cancel, except for the pair that lie on the x axis passing through the center. This pair of charges produces a field pointing to the right which is give n using equation (i) as: 

$\begin{array}{c}\overrightarrow{E}=\frac{3e}{4\pi {\epsilon }_o{d}^2}\widehat{i}\\ =3(8.99\times {10}^9{\text{N.m}}^{\text{2}}{\text{/C}}^{\text{2}})\frac{(1.6\times {10}^{-19}\text{C})}{{\left(0.020\text{m}\right)}^2} \widehat{i}\\ =(1.08\times {10}^{-5}\text{N/C}) \widehat{i}\end{array}$

Hence, the value of the electric field is.$(1.08\times {10}^{-5}\text{N/C}) \widehat{i}$