Q58P - Introduction to Electrodynamics

Question

(a) Consider an equilateral triangle, inscribed in a circle of radius a, with a point charge q at each vertex. The electric field is zero (obviously) at the center, but (surprisingly) there are three other points inside the triangle where the field is zero. Where are they? [Answer: r = 0.285 a-you'll probably need a computer to get it.]

 

(b) For a regular n-sided polygon there are n points (in addition to the center) where the field is zero. Find their distance from the center for n = 4 and n = 5. What do you suppose happens as $\mathit{n}\mathbf{\to }\mathbf{\infty }$?

Step-by-Step Solution

Answer

(a) The electric field of an equilateral triangle, inscribed in a circle of radius a, with a point charge q at each vertex is zero at a distance $0.284718a$from the center.

(b) The electric field of a square of side a, with a point charge q at each vertex is zero at a distance   0.546936a from the center. The electric field of a regular pentagon, inscribed in a circle of radius a, with a point charge q at each vertex is zero at a distance 0.688917a from the center. The distance where the field is zero grows with the number of vertices. For $n\to \infty$ , the points will coincide with the circle of charges. 

Step 1: Given data

(a) An equilateral triangle is inscribed in a circle of radius a, with a point charge q at each vertex.

(b) A square of side a, with a point charge q at each vertex. A regular pentagon is inscribed in a circle of radius a, with a point charge q at each vertex.

Step 2: Electric field of a point charge

The electric field at a distance r from charge q is

  

$\overrightarrow{E}=\frac{1}{4\tau \tau {\epsilon }_0}\frac{q}{r^2}\hat{r} .......\left(1\right)$

Here, ${\epsilon }_0$is the permittivity of free space.

Step 3: Zero field point of equilateral circle

The configuration is shown in the following figure.

From equation (1), the  component of electric field of such a configuration is

$E=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{{(a+r)}^2}-\frac{2 \cos \theta }{b^2}\right]$

  

Thus, the condition for the field to be zero is

$\frac{2 \cos \theta }{b^2}=\frac{1}{{(a+r)}^2} ........\left(2\right)$  

 

But it can be seen that

$\cos \theta =\frac{{\displaystyle \frac{a}{2}}-r}{b}$

and

$b^2={\left(\frac{a}{2}-r\right)}^2+{\left(\frac{\sqrt{3}a}{2}\right)}^2$

  

Substitute these two values in equation (2) and get

  

 $\frac{2\left({\displaystyle \frac{a}{2}}-r\right)}{{(a^2-ar+r^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{1}{{(a+r)}^2}$

Substitute $\frac{r}{a}=u$in the above equation and get

${(1-2u)}^2{(1+u)}^4={(1-u+u^2)}^3$  

 

Multiply both sides and simplify to get

$u(1-4u+u^3+5u^4+u^5)=0$

Here, u = 0 is one solution which represents the center of the circle. The other solution obtained from Mathematica inside the rectangle is

u = 0.284718

r = 0.284718a  

Thus, the field is zero at r = 0.284718a  distance from the center. 

Step 4: Zero field point of square

The configuration is shown in the following figure.

From equation (1), the x component of electric field of such a configuration is

 

$E=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{2 \cos {\theta }_{+}}{b_{+}^2}-\frac{2 \cos {\theta }_{-}}{b_{-}^2}\right]$

 

Thus, the condition for the field to be zero is

 $\frac{2 \cos {\theta }_{+}}{b_{+}^2}=\frac{2 \cos {\theta }_{-}}{b_{-}^2} ......\left(3\right)$ 

 

But it can be seen that

 

$\cos \theta =\frac{{\displaystyle \frac{a}{\sqrt{2}}}\pm r}{b_{\pm }}$

 

and

$$\begin{gathered} b^2={\left(\frac{a}{\sqrt{2}}\right)}^2+{\left(\frac{a}{\sqrt{2}}\pm r\right)}^2 \\ =a^2\pm \sqrt{2}ar+r^2 \end{gathered}$$

Substitute these two values in equation (3) and get

$\frac{{\displaystyle \frac{a}{\sqrt{2}}}+r}{{(a^2+\sqrt{2}ar+r^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{{\displaystyle \frac{a}{\sqrt{2}}-r}}{{(a^2+\sqrt{2}ar+r^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Substitute  $\frac{\sqrt{2}r}{a}=u$in the above equation and get

${(1+u)}^2{(2-2u+u^2)}^3=(1-u{)}^2{(2+2u+u^2)}^3$  

 

Multiply both sides and simplify to get

$8u-16u^3+2u^5+4u^7=0$

The solution obtained from Mathematica inside the square is

 $$\begin{gathered} u^2=0.598279 \\ u=0.546936a \end{gathered}$$

Thus, the field is zero at 0.546936a distance from the center.

Step 5: Zero field point of regular pentagon

The configuration is shown in the following figure.

From equation (1), the  component of electric field of such a configuration is

$E=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{{(a+r)}^2}+\frac{2 \cos \theta }{b^2}-\frac{2 \cos \varphi }{c^2}\right]$

Thus, the condition for the field to be zero is

$\frac{1}{{(a+r)}^2}=\frac{2 \cos \varphi }{c^2}-\frac{2 \cos \theta }{b^2}$  

 

But it can be seen that

$$\begin{gathered} \cos \theta =\frac{a \cos {\displaystyle \frac{2\mathrm{\pi }}{5}}+r}{b} \\ \cos \varphi =\frac{a \cos {\displaystyle \frac{\mathrm{\pi }}{5}}-r}{c} \end{gathered}$$

and

  

$$\begin{gathered} b^2={\left(a \cos \frac{2\mathrm{\pi }}{5}+r\right)}^2+{\left(a \sin \frac{2\mathrm{\pi }}{5}\right)}^2 \\ =a^2+r^2+2ar \cos \frac{2\mathrm{\pi }}{5} \\ {c}^2={\left(a \cos \frac{2\mathrm{\pi }}{5}-r\right)}^2+{\left(a \sin \frac{\mathrm{\pi }}{5}\right)}^2 \\ =a^2+r^2- 2ar \cos \frac{\mathrm{\pi }}{5} \end{gathered}$$

 

Substitute these in equation (4) and get

  

 $\frac{1}{{(a+r)}^2}+2 \frac{a \cos {\displaystyle \frac{2\mathrm{\pi }}{5}}+r}{{\left(a^2+r^2+2 ar \cos {\displaystyle \frac{2\mathrm{\pi }}{5}}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+2\frac{a \cos {\displaystyle \frac{2\mathrm{\pi }}{5}}-r}{{\left(a^2+r^2-2 ar \cos {\displaystyle \frac{2\mathrm{\pi }}{5}}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=0$

The solution obtained from Mathematica inside the pentagon is

r = 0.688917 a  

 

Thus, the field is zero at 0.688917a distance from the center.