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 $-q$ ?

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

Answer 1

Answer: (B)

If the charge $q$ at $A$ is replaced by $-q$, it is equivalent to adding charge $-2q$ at $A$. Therefore, electric field at $O$, $E_2=\frac{2q}{4\pi {\epsilon }_0{r}^2}$ along $OA$.