Question 761: Electrostats & Capacitance

Question

The point charges $+\mathrm{q},-\mathrm{q},-\mathrm{q},+\mathrm{q},+\mathrm{Q}$ 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 twice the electric field at centre ' O ' due to charge +Q at E alone. The value of $Q$ is

Accepted Answer

Electric field at ' $O$ ' due to charge ' $Q$ ' is given by,
$$ \mathrm{E}=\frac{\mathrm{kQ}}{\mathrm{r}^2} $$
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, $\mathrm{E}^{\prime}=2 \mathrm{E}$
$$ 	herefore \quad \frac{\mathrm{kq}}{\mathrm{r}^2}=\frac{2 \mathrm{kQ}}{\mathrm{r}^2} \Rightarrow \mathrm{Q}=\frac{\mathrm{q}}{2} $$

Answer

$\frac{\mathrm{q}}{2}$

Answer

q

Answer

$2 q$

Answer

$4 q$

Electrostats & Capacitance

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