91Ó°ÊÓ

(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

$\stackrel{→}{E}=\frac{1}{4\mathrm{Ï„}\mathrm{Ï„}{\mathrm{Î\mu }}_0}\frac{q}{r^2}\hat{r}.......\left(1\right)$

Here, ${\mathrm{Î\mu }}_0$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

$E=\frac{q}{4{\mathrm{Ï€Î\mu }}_0}\left[\frac{1}{{(a+r)}^2}-\frac{2\cos \mathrm{θ}}{b^2}\right]$

Thus, the condition for the field to be zero is

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

But it can be seen that

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

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

$E=\frac{q}{4{\mathrm{Ï€Î\mu }}_0}\left[\frac{2\cos {\mathrm{θ}}_{+}}{b_{+}^2}-\frac{2\cos {\mathrm{θ}}_{-}}{b_{-}^2}\right]$

Thus, the condition for the field to be zero is

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

But it can be seen that

$\cos \mathrm{θ}=\frac{{\displaystyle \frac{a}{\sqrt{2}}}Â\pm r}{b_{Â\pm }}$

and

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

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

$E=\frac{q}{4{\mathrm{Ï€Î\mu }}_0}\left[\frac{1}{{(a+r)}^2}+\frac{2\cos \mathrm{θ}}{b^2}-\frac{2\cos \mathrm{Ï•}}{c^2}\right]$

Thus, the condition for the field to be zero is

$\frac{1}{{(a+r)}^2}=\frac{2\cos \mathrm{ϕ}}{c^2}-\frac{2\cos \mathrm{θ}}{b^2}$

But it can be seen that

$$\begin{gathered} \cos \mathrm{θ}=\frac{a\cos {\displaystyle \frac{2\mathrm{π}}{5}}+r}{b} \\ \cos \mathrm{ϕ}=\frac{a\cos {\displaystyle \frac{\mathrm{π}}{5}}-r}{c} \end{gathered}$$

and

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

Substitute these in equation (4) and get

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