Four point charges of +100 esu,-50 esu +20 esu and +30 esu are placed respectively at the four vertices A,B,C,D of a square of side 10 cm. Find the electric intensity and potential at the point of intersection of the diagonals.

Explanation

Length of the diagonal of the square
`= sqrt(10^(2) + 10^(2)) = 10sqrt(2) cm`
The point of intersection of the two diagonals is the point O.

`:. AO = BO = CO = DO = (10sqrt2)/2 = 5sqrt(2)cm`
Intensity acts along `vec(OC)` due to the charge at A and it acts along `vec(OA)` due to the charge at C.
So resultant intensity at O due to the charges at A and C is given by
`E_(1) = 100/((5sqrt2)^(2)) - 20/((5sqrt2)^(2))`
`=(2-2/5) = 8/5 dyn cdot statC^(-1) , "along" vec(OC)`
Similarly intensity at O due to the charges at B and D acts along `vec(OB)`.
So, resultant intensity at O due to the charges at B and D is given by,
`E_(2) = 30/((5sqrt2)^(2))+50/((5sqrt2)^(2))`
`= (3/5+1) = 8/5 dyn cdot statC^(-1) , "along" vec(OB)`
Here `E_(1)` and `E_(2)` act perpendicular to each other and `E_(1) = E_(2)`.
`:.` Resultant intensity at O,
`E = sqrt(E_(1)^(2)+E_(2)^(2)) = sqrt((8/5)^(2)+(8/5)^(2)) = 8/5sqrt(2)`
`=2.263 dyn cdot stat C^(-1)`
The direction of E is along the bisector `vec(OP)` of the angle COB, i.e., parallel to the side AB or DC.
Again, potential at `O = 100/(5sqrt2) - 50/(5sqrt2) + 20/(5sqrt2) + 30/(5sqrt2)`
`=100/(5sqrt2) = 10sqrt(2) = 14.14 statV`.