Two point charges of values -20 esu and +20 esu are placed on the x-axis at `x = -10cm and x = +10cm` respectively. Calculate
(ii) The electric fields at the points `P(1,10) and Q(20,0)`.

Explanation

Two charges + 20 esu and -20 esu are placed at A and B respectively.
(ii) Electric field intensity atP due to the charge -20 esu at A is
`E_(1) = 20/(AP^2) `, along `vec(AP)`
Again, electric field intensity at P due to the charge + 20 esu at B is
`E_(1) = 20/(BP^2) `, along extended `vec(BP)`
Along y-axis, component of `E_(1)`
`(E_1)_(y) = 20/(AP_(2))sin theta , "along" vec(PY)`
`:' AP = BP`
`:. (E_1)_(y) = -(E_2)_(y)`
Hence along y-axis, resulant of component of `E_(1)` and `E_(2)` become zero.
Again along x-axis, component of `E_(1)` is
`(E_1)_(x) = 20/(AP^(2)) cos theta , along vec(PE)`
`(E_2)_(x) = 20/(BP^(2)) cos theta , along vec(PE)`
`(E_1)_(x) and (E_2)_(x)` are equal and acting in the same direction.
`:.` Resultant of `(E_1)_(x) and (E_2)_(x)`
`= E_(x) = 20/(AP^2) cos theta + 20/(BP^2) cos theta = 2 xx 20/(AP^2) cos theta`
Thus resultant of y component of electric field intensity is zero, so
`E_(P) = E_(x) = (2 xx 20)/(AP^2) xx (OA)/(AP) = (2xx 20 xx 10)/((10^(2)+10^(2))^(3//2))`
`= 0.14 "dyne"//"stat coulomb", "along" vec(PE)`
And intensity at Q,
`E_(Q) = 20/(10^2) - 20/(30^2) = 20/100 - 20/900 = 8/45`
`=0.178 = 1.18 "dyne"//"stat coulomb", "along" vec(QX)`.