Question

Five charges, $q$ each is placed at the corners of a regular pentagon of side $a$ as shown in figure. What will be the electric field at $O$ if the charge $q$ at A is replaced by $-q$?

Answer 1

When the charge $q$ from $A$ is removed and $-q$ charge is placed then electric field at $O$

$E_2=\frac{-q}{4\pi {ϵ}_0{r}^2}$.

When a new charge $-q$ is placed the $E.F.$ will increase by $\frac{-q}{4\pi {ϵ}_0{r}^2}$.

Now the resultant becomes

$E_3=E_2+\frac{-1q}{4\pi {ϵ}_0{r}^2}=\frac{-q}{4\pi {ϵ}_0{r}^2}+\frac{-q}{4\pi {ϵ}_0{r}^2}$,

$E_3=\frac{-2q}{4\pi {ϵ}_0{r}^2}$