Question

FIGURE shows a thin rod of length $L$with total charge $Q$.

a. Find an expression for the electric field strength at point $P$on the axis of the rod at distance $r$from the center.

b. Verify that your expression has the expected behavior if $r\gg L$.

c. Evaluate $E$at $r=3.0cm$if $L=5.0cm$and $Q=3.0\mathrm{nC}$.

Short answer

Part$\left(a\right)$

$\left(a\right)$The expression for the electrical field, $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2-\frac{L^2}{4}}\hat{\mathbf{i}}.$

Part$\left(b\right)$

$\left(b\right)$The expression reduces to the expression for the point charge, $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}\hat{\mathbf{i}}.$

Part$\left(c\right)$

$\left(c\right)$The field strength is $E=9.8·{10}^4\mathrm{N}/\mathrm{C}.$

Step-by-step solution

Find the expression (part a)

As illustrated in the illustration, move the $x$axis all along rod with origin at the end. The linear charge density is calculated as follows:

$\lambda =\frac{Q}{L}$

It has a charging aspect to it.

$$\begin{gathered} dQ=\lambda dx \\ =\frac{Q}{L}dx \end{gathered}$$

Because the wire section is so small, it can be viewed as a point charge. As a result, the field element produced just at point away from the wire along the axis is given by

$$\begin{gathered} d\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{{\left(r+\frac{L}{2}-x\right)}^2}\hat{\mathbf{i}} \\ =\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L}\frac{dx}{{\left(r+\frac{L}{2}-x\right)}^2}\hat{\mathbf{i}} \end{gathered}$$

The equation is integrated as,

$\overrightarrow{E}={\int }_0^{L}d\overrightarrow{E}$

Remove everything in integral that is constant.

$\overrightarrow{E}=\hat{\mathbf{i}}\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L}{\int }_0^L\frac{dx}{{\left(r+\frac{L}{2}-x\right)}^2}$

Explained by diagram (part a)

Denote,

$Int={\int }_0^L\frac{dx}{{\left(r+\frac{L}{2}-x\right)}^2}$

Introduce a substitution,

$\xi =x-r-\frac{L}{2},d\xi =dx$

The new integration bounds are

$$\begin{gathered} x=0\to \xi =-r-\frac{L}{2}; \\ x=L\to \xi =-r+\frac{L}{2} \end{gathered}$$

$$\begin{gathered} Int={\int }_{-r-\frac{L}{2}}^{-r+\frac{L}{2}}\frac{d\xi }{{\xi }^2} \\ =-{\left.\frac{1}{\xi }\right|}_{-r-\frac{L}{2}}^{-r+\frac{L}{2}} \\ =\frac{1}{r-\frac{L}{2}}-\frac{1}{r+\frac{L}{2}} \\ =\frac{L}{r^2-\frac{L^2}{4}} \end{gathered}$$

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2-\frac{L^2}{4}}\hat{\mathbf{i}}$

The expression reduces to the expression for the point charge (part b)

In this case, $r\gg L$then $\frac{L^2}{4}$can be neglected by $r^2$. so,

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}\hat{\mathbf{i}}$

This turns out to be an expression for the expected charge far away from the point; the entire wire appears to be a point charge.

Field strength (part c)

The vector sum of all forces exerted by a field at a particular position within the field on a unit mass, unit charges, unit magnetic pole, and so on from the figure as

$E=9.8·{10}^4\mathrm{N}/\mathrm{C}.$(constant)