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

(b) For a regular n-sided polygon there are npoints (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 $\mathit{n}\mathbf{\to }\mathbf{\infty }$?

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

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

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.

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.

The configuration is shown in the following figure.

From equation (1), the xcomponent 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

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.

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+2ar\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^2-2ar\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.