Question 807: 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 $A, B, C, D$ and $F$ is thrice 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, $\mathrm{E}^{\prime}=3 \mathrm{E}$
$$	herefore\quad\frac{\mathrm{kq}}{\mathrm{r}^2}=\frac{3 \mathrm{kQ}}{\mathrm{r}^2} \Rightarrow \mathrm{Q}=\frac{\mathrm{q}}{3}$$

Answer

$\frac{+q}{3}$

Answer

$\frac{+q}{5}$

Answer

$\frac{+q}{6}$

Answer

$+6 q$

Electrostats & Capacitance

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