Question

Six charges are placed at the vertices of a regular hexagon as shown in the figure. Find the electric field on the line passing through $O$ and perpendicular to the plane of the figure as a function of distance $x$ from point $O$:

• A. $0$
• B. $4$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$0$

The electric field due to a charge is given by $\overrightarrow{E}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^3}\overrightarrow{r}$

Thus, the electric field at $P$ which is located at a distance $x$ from $O$ is given by superposition of fields due to individual charges.

Let $k=\frac{1}{4\pi {ϵ}_0}$ and $d=\sqrt{a^2+x^2}$

Then ${\overrightarrow{E}}_A=\frac{kQ}{d^3}(x\hat{k}-\overrightarrow{OA})$

Similarly, ${\overrightarrow{E}}_B=\frac{-kQ}{d^3}(x\hat{k}-\overrightarrow{OB})$ , ${\overrightarrow{E}}_C=\frac{kQ}{d^3}(x\hat{k}-\overrightarrow{OC}),$

${\overrightarrow{E}}_D=\frac{-kQ}{d^3}(x\hat{k}-\overrightarrow{OD}),$

${\overrightarrow{E}}_E=\frac{kQ}{d^3}(x\hat{k}-\overrightarrow{OE}),$

${\overrightarrow{E}}_F=\frac{-kQ}{d^3}(x\hat{k}-\overrightarrow{OF})$

Thus, $\overrightarrow{E}=\frac{kQ}{d^3}(x\hat{k}(1-1+1-1+1-1)+(\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OE})-(\overrightarrow{OB}+\overrightarrow{OD}+\overrightarrow{OF}))=0$