5.4 Electric field  (Page 4/7)

Solution

1. By symmetry, the horizontal ( x )-components of $\overrightarrow{\text{E}}$ cancel ( [link] ); $E_x=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\text{sin}\theta -\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\text{sin}\theta =0$ .
   

   

   
   The vertical ( z )-component is given by
   
   $E_z=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\text{cos}\theta +\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\text{cos}\theta =\frac{1}{4\pi {\epsilon }_0}\frac{2q}{r^2}\text{cos}\theta .$
   
   Since none of the other components survive, this is the entire electric field, and it points in the $\widehat{\text{k}}$ direction. Notice that this calculation uses the principle of superposition    ; we calculate the fields of the two charges independently and then add them together.
   What we want to do now is replace the quantities in this expression that we don’t know (such as r ), or can’t easily measure (such as $\text{cos}\theta )$ with quantities that we do know, or can measure. In this case, by geometry,
   
   $r^2=z^2+{\left(\frac{d}{2}\right)}^2$
   
   and
   
   $\text{cos}\theta =\frac{z}{r}=\frac{z}{{\left[z^2+{\left(\frac{d}{2}\right)}^2\right]}^{1\text{/}2}}.$
   
   Thus, substituting,
   
   $\overrightarrow{\text{E}}(z)=\frac{1}{4\pi {\epsilon }_0}\frac{2q}{{\left[z^2+{\left(\frac{d}{2}\right)}^2\right]}^2}\frac{z}{{\left[z^2+{\left(\frac{d}{2}\right)}^2\right]}^{1\text{/}2}}\widehat{\text{k}}.$
   
   Simplifying, the desired answer is
   
   $\overrightarrow{\text{E}}(z)=\frac{1}{4\pi {\epsilon }_0}\frac{2qz}{{\left[z^2+{\left(\frac{d}{2}\right)}^2\right]}^{3\text{/}2}}\widehat{\text{k}}.$
2. If the source charges are equal and opposite, the vertical components cancel because $E_z=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\text{cos}\theta -\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\text{cos}\theta =0$
   and we get, for the horizontal component of $\overrightarrow{\text{E}}$ ,
   
   $\begin{array}{cc}\hfill \overrightarrow{\text{E}}(z)& =\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\text{sin}\theta \widehat{\text{i}}-\frac{1}{4\pi {\epsilon }_0}\frac{\text{−}q}{r^2}\text{sin}\theta \widehat{\text{i}}\hfill \\ & =\frac{1}{4\pi {\epsilon }_0}\frac{2q}{r^2}\text{sin}\theta \widehat{\text{i}}\hfill \\ & =\frac{1}{4\pi {\epsilon }_0}\frac{2q}{{\left[z^2+{\left(\frac{d}{2}\right)}^2\right]}^2}\frac{\left(\frac{d}{2}\right)}{{\left[z^2+{\left(\frac{d}{2}\right)}^2\right]}^{1\text{/}2}}\widehat{\text{i}}.\hfill \end{array}$
   
   This becomes
   
   $\overrightarrow{\text{E}}(z)=\frac{1}{4\pi {\epsilon }_0}\frac{qd}{{\left[z^2+{\left(\frac{d}{2}\right)}^2\right]}^{3\text{/}2}}\widehat{\text{i}}.$

Significance

Let’s start with [link] , the field of two identical charges. From far away (i.e., $z\gg d),$ the two source charges should “merge” and we should then “see” the field of just one charge, of size 2 q . So, let $z\gg d;$ then we can neglect $d^2$ in [link] to obtain

which is the correct expression for a field at a distance z away from a charge 2 q .

Next, we consider the field of equal and opposite charges, [link] . It can be shown (via a Taylor expansion) that for $d\ll z\ll \infty$ , this becomes

which is the field of a dipole, a system that we will study in more detail later. (Note that the units of $\overrightarrow{\text{E}}$ are still correct in this expression, since the units of d in the numerator cancel the unit of the “extra” z in the denominator.) If z is very large $\left(z\to \infty \right)$ , then $E\to 0$ , as it should; the two charges “merge” and so cancel out.

Summary

• The electric field is an alteration of space caused by the presence of an electric charge. The electric field mediates the electric force between a source charge and a test charge.
• The electric field, like the electric force, obeys the superposition principle
• The field is a vector; by definition, it points away from positive charges and toward negative charges.

Conceptual questions