Question

Five equal and similar charges are placed at the corners of a regular hexagon as shown in figure. What is the electric field and potential at the centre of the hexagon?

• A. $\frac{5}{3\pi {\epsilon }_0}\frac{q}{l},\frac{5}{4\pi {\epsilon }_0}\frac{q}{l^2}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{q}{l},\frac{5}{4\pi {\epsilon }_0}\frac{q}{l^2}$
• C. $\frac{5}{4\pi {\epsilon }_0}\frac{q}{l},\frac{1}{4\pi {\epsilon }_0}\frac{q}{l^2}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{q}{l},\frac{1}{4\pi {\epsilon }_0}\frac{q}{l^2}$

Answer 1

Answer: (C)

$\frac{5}{4\pi {\epsilon }_0}\frac{q}{l},\frac{1}{4\pi {\epsilon }_0}\frac{q}{l^2}$

Here the distances from centre O to each charge is $r=\frac{l/2}{\sin 30}=l$.

As the potential is a scalar quantity so the potential at O is

$V=\frac{1}{4\pi {ϵ}_0}[q/r+q/r+q/r+q/r+q/r]=\frac{5}{4\pi {ϵ}_{0}l}$ as $r=l$

As the electric field is a vector quantity so field at O cancel each other due to the charges at opposite corner and only one charge will contribute the field.

Thus, $E=\frac{q}{4\pi {ϵ}_0{r}^2}=\frac{q}{4\pi {ϵ}_0{l}^2}$