The distance between two points A and B in vacuum is 2d. At each of these two points a + Q charge is placed. P is the midpoint of AB. Find the intensity and potential at P due to the electric field. How will the values of these quantities change if the charge at B is replaced by a charge -Q ?

Explanation

`AB - 2d, "mid point of" AB is P, i.e., AP = BP = d`

Intensity at P due to the charge +Q at A
`Q/(d^2)`, along `vec(PB)` …………(1)
Intensity at P due to the charge + Q at B
`=Q/(d^2), "along" vec(PA)` ...............(2)
So, intensities at P due to the charges at A and B are equal and opposite.
Therefore, the resultant intensity at P is zero.
Potential at P due to the charge at `A = Q/d` ..........(3)
Potential at P due to the charges at `B = Q/d` ..............(4)
`:.` Total potential at `P = Q/d + Q/d = (2Q)/d`.
Now if -Q charge be placed at the point B, intensity in equation (2) Will be `Q/ (d^2)` along `vec(PB)`
`:.` Resultant intensity at `P = Q/(d^2) + Q/(d^2) = (2Q)/(d^2)`, along `vec(PB)`
Again the value of potential in equation (4) will be `(-Q)/(d)`.
`:.` Total potential at `P = Q/d - Q/d = 0`.