Question

(a) Find thecomponents of the electric field at the point P on the y axis a distance d from the origin.

Ex	=
Ey

(b) What are the approximate values of the field components when d >> L?

Answer 1

The other poster made some major errors. The solution below is completely correct.

The charge per unit length of the rod is:

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

Any small element of the rod a distance \(x\) from the origin has a charge of:

dl \(\mathbf{q}=\sigma d \mathbf{x}=\frac{Q}{L} d \mathbf{x}\)

The distance. \(r\), between \(P\) and that element located a distance \(x\) from

the origin is:

$$ x=\sqrt{d^{2}+x^{2}} $$

The unit direction of the electric field at point \(P\) from that one element of the rod a distance \(x\) from the origin is:

$$ \vec{\lambda}=\frac{\{-x, d\}}{r}=\frac{\{-x, d\}}{\sqrt{d^{2}+x^{2}}} $$

Thus the electric field at \(P\) from that one element from Coulomb's law

is:

\(\mathrm{d} \overrightarrow{\mathrm{E}}=\) ke \(\frac{\mathrm{d} \mathrm{q}}{r^{2}} \vec{\lambda}\)

d\(\overrightarrow{\mathbf{E}}=\operatorname{ke} \frac{\mathrm{Q}}{\mathrm{L}}\left\{-\frac{\mathrm{x}}{\left(\mathrm{d}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}, \frac{\mathrm{d}}{\left(\mathrm{d}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}\right\} \mathrm{dx}\)

The total electric field at \(P\) from the rod is the sum of every \(d \vec{E}\) from

\(x=0\) to \(x=L\)

\(\overrightarrow{\mathrm{E}}=\operatorname{ke} \frac{\mathrm{Q}}{\mathrm{L}} \int_{0}^{\mathrm{L}}\left\{-\frac{\mathrm{x}}{\left(\mathrm{d}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}, \frac{\mathrm{d}}{\left(\mathrm{d}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}\right\} \mathrm{dx}\)

\(\overrightarrow{\mathrm{E}}=\operatorname{ke} \frac{\mathrm{Q}}{\mathrm{L}}\left\{\int_{0}^{\mathrm{L}}-\frac{\mathrm{x}}{\left(\mathrm{d}^{2}+\mathrm{x}^{2}\right)^{3 / 2}} \mathrm{dx}, \int_{0}^{\mathrm{L}} \frac{\mathrm{d}}{\left(\mathrm{d}^{2}+\mathrm{x}^{2}\right)^{3 / 2}} \mathrm{dx}\right\}\)

\(\overrightarrow{\mathrm{E}}=\operatorname{ke} \frac{\mathrm{Q}}{\mathrm{L}}\left\{\left(\frac{1}{\sqrt{\mathrm{d}^{2}+\mathrm{x}^{2}}}\right)_{0}^{\mathrm{L}},\left(\frac{\mathrm{x}}{\mathrm{d} \sqrt{\mathrm{d}^{2}+\mathrm{x}^{2}}}\right)_{0}^{\mathrm{L}}\right\}\)

\(\overrightarrow{\mathrm{E}}=\operatorname{ke} \frac{\mathrm{Q}}{\mathrm{L}}\left\{\frac{1}{\sqrt{\mathrm{d}^{2}+\mathrm{L}^{2}}}-\frac{1}{\mathrm{~d}}, \frac{\mathrm{L}}{\mathrm{d} \sqrt{\mathrm{d}^{2}+\mathrm{L}^{2}}}\right\}\)

And breaking into components yields:

\(\mathrm{Ex}=\operatorname{ke} \frac{\mathrm{Q}}{\mathrm{L}}\left(\frac{1}{\sqrt{\mathrm{d}^{2}+\mathrm{L}^{2}}}-\frac{1}{\mathrm{~d}}\right)\)

\(E y=\) ke \(Q \frac{1}{d \sqrt{d^{2}+L^{2}}}\)

When \(\mathrm{d}>>\mathrm{L}\), then we omit \(\mathrm{L}\) when it is grouped with \(\mathrm{d}\) since \(d^{2}+L^{2} \simeq d^{2}:\)

\(\overrightarrow{\mathrm{E}}=\operatorname{ke} \frac{\mathrm{Q}}{\mathrm{L}}\left\{\frac{1}{\mathrm{~d}}-\frac{1}{\mathrm{~d}}, \frac{\mathrm{L}}{\mathrm{d}^{2}}\right\}\)

\(E x=0\)

\(\mathrm{Ey}=\frac{\text { ke } \mathrm{Q}}{\mathrm{d}^{2}}\)