Question

Five point charges, each of value +q, are placed on five vertices of a regular hexagon of side L. The magnitude of the force on a point charge of value –q placed at the centre of the hexagon (in newton) is

• A. Zero
• B. $\frac{\sqrt{3}{q}^2}{4\pi {ϵ}_0{L}^2}$
• C. $\frac{q^2}{4\pi {ϵ}_0{L}^2}$
• D. $\frac{q^2}{4\sqrt{3}\pi {ϵ}_0{L}^2}$

Answer 1

The correct option is C $\frac{q^2}{4\pi {ϵ}_0{L}^2}$

Suppose six charges were placed at the corners 1,2,3,4,5 and 6 of hexagon would have been zero.

$Rightarrow\stackrel{\rightharpoondown }{F_{01}}+\stackrel{\rightharpoondown }{F_{02}}+\stackrel{\rightharpoondown }{F_{03}}+\stackrel{\rightharpoondown }{F_{04}}+\stackrel{\rightharpoondown }{F_{05}}+\stackrel{\rightharpoondown }{F_{06}}=0$
$\Rightarrow \stackrel{\rightharpoondown }{F_{01}}+\stackrel{\rightharpoondown }{F_{02}}+\stackrel{\rightharpoondown }{F_{03}}+\stackrel{\rightharpoondown }{F_{04}}+\stackrel{\rightharpoondown }{F_{05}}=-\stackrel{\rightharpoondown }{F_{06}}$
$\Rightarrow \left|\begin{array}{c}Resultantofforcesactingoncharge\\ placedatthecentreOofhexagon\end{array}\right|=|\stackrel{\rightharpoondown }{F_{06}}|$
$\Rightarrow |Resultant|=\frac{q^2}{4\pi {ϵ}_0{r}^2};$ since $r=L\{\because sin{30}^{\circ }=\frac{L}{2r}\}$
$|Resultant|=\frac{q^2}{4\pi {ϵ}_0{r}^2}$