Five identical charges Q are placed equidistant on a semicircle as shown in the figure . Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Explanation

Force acting on q due to `Q_(1)` and `Q_(5)` are opposite direction so cancel to each other.

Force acting on q due to `Q_(3)` is `F_(3)= (1)/(4piepsilon_(0))(qQ_(3))/(R^(2))`
Force acting on q due to `Q_(2)` and `Q_(4)`
Resolving in two component method:
(i) Vertical component : `Q_(2) " Sin" theta "and " Q_(4) " Sin" theta` are equal and opposite direction so they are cancel to each other.
(ii) Horizontal Component : `Q_(2) "Sin" theta "and " Q_(4) "cos " theta ` are equal and same direction so they can get added.
`F_(24)=F_(2q)+F_(4q)= F_(2) "cos " 45^(@)+ F_(4)"cos"45^(@)`
`F_(24)= (1)/(4piepsilon_(0))(qQ_(2))/(R^(2))"cos" 45^(@)+ (1)/(4piepsilon_(0)) (qQ4)/(R^(2))"cos" 45^(@)`
Resultant net force F `= F_(3)+F_(24)+F_(15)`
`Q=Q_(1)=Q_(2)=Q_(3)=Q_(4)=Q_(5)`
`"cos" 45^(@) = (1)/(sqrt(2))`
`F=(qQ)/(4piepsilon_(0)R^(2))[1+(1)/(sqrt(2))+(1)/(sqrt(2))]`
`=(1)/(4piepsilon_(0))(qQ)/(R^(2))[1+(2)/(sqrt(2))]`
`F= (1)/(4piepsilon_(0))(qQ)/(R^(2))[1+sqrt(2)]N`
Vector form :
`vecF=(1)/(4piepsilon_(0))(qQ)/(R^(2))(1+sqrt(2))N hati`