## In This post you will get answer of Q.

Four point charges (with equal magnitude of charge of $5, C$; but with different signs) are placed at four corners of a square of side $10, m.$ Assuming that the square is centered at the origin and the configuration of the charges are as given in the figure, the potential and the magnitude of electric field at the origin, respectively are [ Note : $k = frac{1}{4 pi varepsilon_0}$]

Solution:

## Four point charges are placed at four corners of a square of side 10 m, is given in figure below as

Now, the potential at point $O$

$V_{ O } =V_{-5 C }+V_{-5 C }+V_{+5 C }+V_{+5 C } $

$=2 V_{-5 C }+2 V_{+5 C } $

$=frac{2}{4 pi varepsilon_{0}} frac{(-5)}{5 sqrt{2}}+frac{2}{4 pi varepsilon_{0}} frac{(+5)}{5 sqrt{2}} $

$=0, V $( zero )

and now electric field at origin

$E_{O B} =frac{1}{4 pi varepsilon_{0}} frac{5}{(5 sqrt{2})^{2}}$

$=k frac{1}{10}$( towards $O$ )

Similarly, $ E_{O C} =k frac{5}{(5 sqrt{2})^{2}}=k frac{1}{10}$ (towards $O$)

$E_{O A} =k frac{1}{10}$( away from $O$)

$E_{O D} =k frac{1}{10}$( away from $O$ )

Now, the vector sum of all the fields

$ E^{2}=left(2 E_{O A}right)^{2}+left(2 E_{O B}right)^{2} $

$=frac{2}{25} k^{2} $

$ Rightarrow E =frac{sqrt{2}}{5} k, V / m $