Q2P - Introduction to Electrodynamics

Question

Find the electric field (magnitude and direction) a distance z above the midpoint between equal and opposite charges ( $\mathbf{+}\mathit{q}$ ), a distanced apart (same as Example 2.1, except that the charge at $\mathit{x}\mathbf{=}\mathbf{+}\frac{\mathbf{d}}{\mathbf{2}}\mathbf{is}\mathbf{-}\mathit{q}\mathbf{)}$.

Step-by-Step Solution

Answer

Two charge $q_1$ and $q_2$, which are equal and opposite in nature, are placed at a distance apart. Then electric force exerted by $q_1$ and $q_2$ at a distance z on the z-axis, at a point above the midpoint of the line joining two charges has to be evaluated.

Step 1: Describe the given information

The distance above the two equal and opposite charges is z.

The two charges are distance d apart.

The charge $-q \mathrm{is} \mathrm{at} \mathrm{x}=\frac{+\mathrm{d}}{2}$.

Step 2: Define the coulombs law

Electric force exerted by charge $\mathit{q}$ on charge $\mathit{Q}$ is proportional to the product of the two charge and inversely proportional to the square of the distance between them as,

$\mathit{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{Q}}{{\mathbf{r}}^{\mathbf{2}}}$

Step 3: Draw the diagram and show the components

Two charge $q_1 \mathrm{and} q_2$, which are equal and opposite in nature, are placed at a distance $d$apart. Then electric force exerted by $q_1 \mathrm{and} q_2$ at a distance z on the z-axis is shown in following figure

                

From the above figure, from the right triangle, using Pythagoras theorem we can write

$r^2=z^2+\frac{d^2}{4}$                  …… (1)

$\frac{d}{2}=r \sin \theta$                   …… (2)

The component of electric field $E_1$ is due to the charge $-q$a, the force is attractive and component of electric field $E_2$ is due to the charge$+q$, the force is repulsive

These electric field can be resolved into two perpendicular components, then the components along the z axis can be cancelled as these are equal and opposite.

Step 4: Find the expression of resultant electric field

The electric field at the point, lying on the line bisecting the line joining two charges at a distance z, is equal to the sum of horizontal components $E_{1x}$and $E_{2x}$as

$$\begin{gathered} E=E_{1x}+E_{2x} \\ =\frac{q\sin \theta }{4\pi {\epsilon }_0}\left(\frac{1}{r^2}\right)+\frac{q\sin \theta }{4\pi {\epsilon }_0}\left(\frac{1}{r^2}\right) \\ =\frac{2q\sin \theta }{4\pi {\epsilon }_0}\left(\frac{1}{r^2}\right) \end{gathered}$$$E_{2x}$

Step 5: Simplify further

From equation (2) we can rearrange it as 

$\sin \theta \frac{\left({\displaystyle \frac{d}{2}}\right)}{r}$                     …… (3)

Substitute $\sqrt{z^2+\frac{d^2}{4}}$ for $r$into equation (3)

$$\begin{gathered} \sin \theta =\frac{\left({\displaystyle \frac{d}{2}}\right)}{\sqrt{z^2+\frac{d^2}{4}}} \\ =\frac{d}{2\left(\sqrt{z^2+\frac{d^2}{4}}\right)} \end{gathered}$$

Substitute $\frac{d}{2\left(\sqrt{z^2+\frac{d^2}{4}}\right)} \mathrm{for} \sin \theta , \mathrm{and} , \sqrt{z^2+\frac{d^2}{4}} \mathrm{for} \mathrm{r} \mathrm{into} \mathrm{E}\frac{2\mathrm{qsin\theta }}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{{\mathrm{r}}^2}\right)$

$$\begin{gathered} E=\frac{2q}{4\pi {\epsilon }_0}\left(\frac{d}{2\left(\sqrt{z^2+\frac{d^2}{4}}\right)}\right)\left(\frac{1}{z^2+{\displaystyle \frac{d^2}{4}}}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{qd}{{\left(z^2+\frac{d^2}{4}\right)}^{3/2}}\right) \end{gathered}$$

The resultant electric field is in x direction. So, the resultant electric field vector can be written as

$E=\frac{1}{4\pi {\epsilon }_0}\left(\frac{qd}{{\left(z^2+\frac{d^2}{4}\right)}^{3/2}}\right)\stackrel{\lambda }{x}$

Therefore, the net or resultant electric field at a distance z above the line midpoint of the line joining two charges is $E=\frac{1}{4\pi {\epsilon }_0}\left(\frac{qd}{{\left(z^2+\frac{d^2}{4}\right)}^{3/2}}\right)\stackrel{\lambda }{x}$.