Question

Five equal charges each of value $q$ are placed at the corners of a regular pentagon of side $a$. What will be the electric field at $O$ if the charge $q$ at $A$ is replaced by $-q$ then?

• A. $\frac{2q}{4\pi {\epsilon }_0{r}^2}$ along $OA$
• B. $\frac{2q}{4\pi {\epsilon }_0{r}^2}$ along $OB$
• C. $\frac{4q}{4\pi {\epsilon }_0{r}^2}$ along $OC$
• D. zero

Answer 1

The correct option is A $\frac{2q}{4\pi {\epsilon }_0{r}^2}$ along $OA$

If all charge will be at the corner then there will not any electric field at the center, because of arrangement is symmetric about the center of the pentagon.

But when one charge removes then equilibrium will disturb and the electric field will be generated toward that vacant corner, and its magnitude will be equal to the -q charge at a point.

but, when another $-q$ charge is kept at that point then

${\overrightarrow{E}}_{-q}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$ along $OA$.

Hence net electric field at the centre

${\overrightarrow{E}}_{net}={\overrightarrow{E}}_{-q}+{\overrightarrow{E}}_{fourcharges}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$

${\overrightarrow{E}}_{net}=\frac{1}{4\pi {\epsilon }_0}\frac{2q}{r^2}$ along $OA$.