Two point charges `q_(1)` and `q_(2)` of magnitude + `10^(-8)` c and `-10^(-8)` c respectively are placed 0.1 m apart calculate the electric fields at points a, b and c

Explanation

We know that direction of electric field due to a positive charge is always away from it and that due to a negative charge is always towards it.

Magnitude of electric field at point A due to q,
`E_(1A) = (kq_(1))/(r_(1A)^(2)) =(9 xx 10^(9))(10^(-8))/(0.05^(2)) = 3.6 xx 10^(4)` N/C
(Direction: From `q_(1)` to `q_(2)`)
Similarly, magnitude of electric field at point A due to `q_(2)`
`E_(2A) = (kq_(2))/r_(2A)^(2) = (9 xx 10^(9))(10^(-6))/(0.5)^(2) = 3.6 xx 10^(4) N//C`
(Direction: From `q_(1)` to `q_(2)`)
If resultant electric field at point A is:
`vecE_(A) = vecE_(1A) + vecE_(2A)`
`therefore E_(A) = E_(1A) + E_(2A) (therefore vecE_(1A) || vecE_(2A))`
`=2E_(1A) (therefore E_(1A) || E_(2A))`
`=2 xx 3.6 xx 10^(4)`
`therefore E_(A) = 7.2 xx 10^(4)` N/C
Magnitude of electric fields at point C due to `q_(1)` and `q_2` are equal which is,
`E = E_(1C) = E_(2C) = (kq_(1))/r_(1C)^(2) = (kq_(2))/(r_(2C)^(2))`
`=(9 xx 10^(9))(10^(-8))/(0.1)^(2) = 9 xx 10^(3) N//C`
Resultant electric field at point C is,
`vecE_(C) = vecE_(1C) + vecE_(2C)`
`therefore E_( C) = sqrt(2E^(2) + 2E^(2) (-1/2)) = sqrt(E^(2)) = E`
`therefore E_( C) = 9 xx 10^(3)` N/C
(In a direction, parallel to direction from `q_(1)` to `q_2`) (Towards right)