(a) One such point is on the x axis (see diagram) at x = r. Here the field is
E_{x}=\frac{q}{4 \pi \epsilon_{0}}\left[\frac{1}{(a+r)^{2}}-2 \frac{\cos \theta}{b^{2}}\right]=0, \quad \text { or } \quad \frac{2 \cos \theta}{b^{2}}=\frac{1}{(a+r)^{2}}.
Now,
\cos \theta=\frac{(a / 2)-r}{b} ; \quad b^{2}=\left(\frac{a}{2}-r\right)^{2}+\left(\frac{\sqrt{3}}{2} a\right)^{2}=\left(a^{2}-a r+r^{2}\right) .
Therefore
\frac{2[(a / 2)-r]}{\left(a^{2}-a r+r^{2}\right)^{3 / 2}}=\frac{1}{(a+r)^{2}} . \quad \text { To simplify, let } \frac{r}{a} \equiv u:
\frac{(1-2 u)}{\left(1-u+u^{2}\right)^{3 / 2}}=\frac{1}{(1+u)^{2}}, \quad \text { or } \quad(1-2 u)^{2}(1+u)^{4}=\left(1-u+u^{2}\right)^{3} .
Multiplying out each side:
1-6 u^{2}-4 u^{3}+9 u^{4}+12 u^{5}+4 u^{6}=1-3 u+6 u^{2}-7 u^{3}+6 u^{4}-3 u^{5}+u^{6} ,
3 u-12 u^{2}+3 u^{3}+3 u^{4}+15 u^{5}+3 u^{6}=0 .
u = 0 is a solution (of course—the center of the triangle); factoring out 3u we are left with a quintic equation:
1-4 u+u^{2}+u^{3}+5 u^{4}+u^{5}=0.
According to Mathematica, this has two complex roots, and one negative root. The two remaining solutions are u = 0.284718 and u = 0.626691. The latter is outside the triangle, and clearly spurious. So r = 0.284718 a. (The other two places where E = 0 are at the symmetrically located points, of course.)
(b) For the square:
E_{x}=\frac{q}{4 \pi \epsilon_{0}}\left(2 \frac{\cos \theta_{+}}{b_{+}^{2}}-2 \frac{\cos \theta_{-}}{b_{-}^{2}}\right)=0 \quad \Rightarrow \quad \frac{\cos \theta_{+}}{b_{+}^{2}}=\frac{\cos \theta_{-}}{b_{-}^{2}} ,
where
\cos \theta_{\pm}=\frac{(a / \sqrt{2}) \pm r}{b_{\pm}} ; \quad b_{\pm}^{2}=\left(\frac{a}{\sqrt{2}}\right)^{2}+\left(\frac{a}{\sqrt{2}} \pm r\right)^{2}=a^{2} \pm \sqrt{2} a r+r^{2} .
Thus
\frac{(a / \sqrt{2})+r}{\left(a^{2}+\sqrt{2} a r+r^{2}\right)^{3 / 2}}=\frac{(a / \sqrt{2})-r}{\left(a^{2}-\sqrt{2} a r+r^{2}\right)^{3 / 2}} .
\text { To simplify, let } w \equiv \sqrt{2} r / a ; \text { then }
\frac{1+w}{\left(2+2 w+w^{2}\right)^{3 / 2}}=\frac{1-w}{\left(2-2 w+w^{2}\right)^{3 / 2}}, \quad \text { or } \quad(1+w)^{2}\left(2-2 w+w^{2}\right)^{3}=(1-w)^{2}\left(2+2 w+w^{2}\right)^{3}.
Multiplying out the left side:
8-8 w-4 w^{2}+16 w^{3}-10 w^{4}-2 w^{5}+7 w^{6}-4 w^{7}+w^{8}=(\text { same thing with } w \rightarrow-w) .
The even powers cancel, leaving
8 w-16 w^{3}+2 w^{5}+4 w^{7}=0, \quad \text { or } \quad 4-8 v+v^{2}+2 v^{3}=0 ,
where v \equiv w^{2} . According to Mathematica, this cubic equation has one negative root, one root that is spurious (the point lies outside the square), and v = 0.598279, which yields
r=\sqrt{\frac{v}{2}} a=0.546936 a .
For the pentagon:
E_{x}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{(a+r)^{2}}+2 \frac{\cos \theta}{b^{2}}-2 \frac{\cos \phi}{c^{2}}\right)=0 ,
where
\cos \theta=\frac{a \cos (2 \pi / 5)+r}{b}, \quad \cos \phi=\frac{a \cos (\pi / 5)-r}{c} ;
b^{2}=[a \cos (2 \pi / 5)+r]^{2}+[a \sin (2 \pi / 5)]^{2}=a^{2}+r^{2}+2 a r \cos (2 \pi / 5) ,
c^{2}=[a \cos (\pi / 5)-r]^{2}+[a \sin (\pi / 5)]^{2}=a^{2}+r^{2}-2 a r \cos (\pi / 5) .
\frac{1}{(a+r)^{2}}+2 \frac{r+a \cos (2 \pi / 5)}{\left[a^{2}+r^{2}+2 a r \cos (2 \pi / 5)\right]^{3 / 2}}+2 \frac{r-a \cos (\pi / 5)}{\left[a^{2}+r^{2}-2 a r \cos (\pi / 5)\right]^{3 / 2}}=0.
Mathematica gives the solution r = 0.688917 a
For an n-sided regular polygon there are evidently n such points, lying on the radial spokes that bisect the sides; their distance from the center appears to grow monotonically with n: r(3) = 0.285, r(4) = 0.547, r(5) = 0.689, . . . . As n →∞ they fill out a circle that (in the limit) coincides with the ring of charge itself
