Five equal charges (each $\mathrm{q}$)are placed at the corners of a regular pentagon of side $\mathrm{a}$. What will be the electric field at $\mathrm{O}$, if the charge $\mathrm{q}$ at $\mathrm{A}$ is replaced by $-\mathrm{q}$?

Five charges, $q$ each are placed at the corners of a regular pentagon of side 'a'.

(i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed?
(iii) What will be the electric field at O if the charge $q$ at A is replaced by $-q$?

How would your answer to be affected if pentagon is replaced by n-sided regular polygon with charge $\mathrm{q}$ at each of its corners?

Two charges $-\mathrm{q}$ each are fixed separated by distance $2\mathrm{d}$. A third charge $\mathrm{q}$ of mass $\mathrm{m}$ placed at the mid-point is displaced slightly by $\mathrm{x}(\mathrm{x}<<\mathrm{d})$ perpendicular to the line joining the two fixed charged as shown in Figure. Show that $\mathrm{q}$ will perform simple harmonic oscillation of time period.

$\mathrm{T}={\left[\frac{8{\mathrm{\pi }}^3{\mathrm{\epsilon }}_0{\mathrm{md}}^3}{{\mathrm{q}}^2}\right]}^{1/2}$