Two point charges `0.01 mu C` and `-0.01 mu C ` are placed 10 cm apart in vacuum . Calculate the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction.

Explanation

According to the question , `q_(1) = 0.01 mu C " " q_(2) = 0.01 mu C`

Magnitude of electric field intensity at mid point P is `E_(p) = sum_(i=1)^(2) (1)/(4pi epsi_(0)) * (q_(1))/(r_(1)^(2))`
or `E = (1)/(4pi epsi_(0)) ((0.01 xx 10^(-6)))/((5 xx 10^(-2))^(2)) + (1)/(4 pi epsi_(0)) ((0.01 xx 10^(-6)))/((5 xx 10^(-2))^(2))`
`E = (2)/(4 pi epsi_(0)) [ (0.01 xx 10^(-6))/(25 xx 10^(-4))] = 7.2 xx 10^(4)` N/C
The direction of electric field is always positive to negative , so that direction of electric field will in direction of PB .