Find the net electric force on charge `q` in the following cases , take `(1)/(4 pi in_(0)) (Qq)/(d^(2)) = F_(0)`.
(a)
(e) The point charges `Q , 80 Q , 27 Q ,... , 1000 Q` are placed at `x = d , 2d , 3d , .... , 10 d`. A point charge `q` is placed at origin.

Explanation

Electric force on `q`:
Due to `4 Q , F_(1) = (1)/( 4 pi in_(0)) (4 Qq)/((2d)^(2)) = F_(0)`, towards right
Due to `2 Q, F_(2) = (1)/( 4 pi in_(0)) (2 Qq)/(d^(2)) = 2 F_(0)` , towards right
Due to `Q , F_(3) = (1)/( 4 pi in_(0)) (Qq)/(q^(2)) = F_(0)` , towards left
Resultant force on `q`:
`F = F_(1) + F_(2) - F_(3) = F_(0) + 2 F_(0) - F_(0) = 2 F_(0)` , towards right
(b) Electric force on `q` :
Due to `9 Q, F_(1) = (1)/( 4 pi in_(0)) (9 Qq)/((3d)^(2)) = F_(0)`, along `AB`
Due to `16 Q , F_(2) = (1)/(4 pi in_(0)) (16 Qq)/((4 d)^(2)) = F_(0)`, along `CB`
Resultant force on `q`:
`F = sqrt(F_(0)^(2) + F_(0)^(2)) = sqrt(2 F_(0))`
`tan theta = (F_(0))/(F_(0)) rArr theta = 45^(@)`
(c) Electric force on `q`:
Due to `Q at B , F_(1) = (1)/(4 pi in_(0)) (Qq)/(d^(2)) = F_(0) , along BA`
Due to `Q at C , F_(2) = (1)/(4 pi in_(0)) (Qq)/(d^(2)) = F_(0) , along CA`
Resultant force on `q`:
(d) Electric force on `q` :
Due to `2Q "at A" , F_(1) = (1)/(4 pi in_(0)) (2Qq)/((sqrt(2)d)^(2)) = F_(0)`, along AB
Due to `2Q at C,F_(2) = (1)/(4 pi in_(0)) (2Qq)/((sqrt(2)d)^(2)) = F_(0)` , along `CB`
Due to `4 Q` at D , `F_(3) = (1)/(4 pi in_(0)) (4 Qq)/((2d)^(2)) = F_(0)` ,along `DB`
Resultant force on `q`:
Resolving forces along `x` and `y` axes
`F_(x) = F_(0) + F_(0) cos 45^(@) = F_(0) (1 + (1)/(sqrt(2)))`
`F_(y) = F_(0) + F_(0) sin 45^(@) = F_(0) (1 + (1)/(sqrt(2)))`
Resultant force is given as
`F = sqrt(F_(x)^(2) + F_(y)^(2)) = sqrt(2) F_(0) (1 + (1)/(sqrt(2)))`
`= F_(0) (sqrt(2) + 1)`
OR
The resultant of two perpendicular forces `F_(0)` will be `sqrt(2) F_(0)` along the third force `F_(0)`. Hence , resultant of three forces is `(sqrt(2) + 1)F_(0)`.
(e ) The electric force on `q` due to charges `Q , 8 Q , 27 Q , .... 1000 Q` will be towards left.
Force on `q`:
Due to `Q , F_(1) = (1)/(4 pi in_(0)) (Qq)/(d^(2)) = F_(0)`
Due to `8Q , F_(2) = (1)/(4 pi in_(0)) (8 Qq)/((2d)^(2)) = 2F_(0)`
Due to `27 Q , F_(3) = (1)/(4 pi in_(0)) (27 Qq)/((3 d)^(2)) = 3 F_(0)`
Due to `1000 Q , F_(10) = (1)/(4 pi in_(0)) (1000 Qq)/((10 d))^(2) = 10 F_(0)`
Net force on `q`:
`F = F_(1) + F_(2) + F_(3) + ... + F_(10)`
`= F_(0) ( 1 + 2 + 3 + ...+ 10)`
`= 55 F_(0)`