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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. |
Answer» Solution :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` |
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