The point charges $+q,-q,-q,+q,+Q$ and $-q$ are placed at the vertices of a regular hexagon ABCDEF as shown…
The point charges $+q,-q,-q,+q,+Q$ and $-q$ are placed at the vertices of a regular hexagon ABCDEF as shown in figure. The electric field at the centre of hexagon ' $O$ ' due to the five charges at $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ and F is thrice the electric field at centre ' O ' due to charge +Q at E alone. The value of Q is
$\frac{+q}{3}$
$\frac{\mathrm{qq}}{5}$
$\frac{+q}{6}$
$+6 q$.
Solution
Electric field at ' $O$ ' due to charge ' $Q$ ' is given by, $E=\frac{k Q}{r^2}$
Electric field at ' $O$ ' due to charges $+q$ at $A$ and $D$ are equal and opposite and hence they will cancel each other. Similarly electric field at ' O ' due to charges -q at F and C will also cancel each other.
Electric field at ' O ' due to charge -q at B is given by,
$E^{\prime}=\frac{k q}{r^2}$ Now, $\mathrm{E}^{\prime}=3 \mathrm{E}$
$\therefore \quad \frac{\mathrm{kq}}{\mathrm{r}^2}=\frac{3 \mathrm{kQ}}{\mathrm{r}^2} \Rightarrow \mathrm{Q}=\frac{\mathrm{q}}{3}$
