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 centre $O$, if the charge from one of the corners (say $A$) is removed?

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

Answer 1

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

When charge q is removed from any point of a regular polygon, it is similar to replacing a $-q$ charge at the same point.

Now, the electric field at the center due to the five charges will be zero whereas the field due to one negative charge will be given as:

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

It means the direction will be away from the center.