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

The forces acting on q, due to `Q_(1)andQ_(2)` are `F_(1)andF_(2)`.

These forces are equal and opposite.
Forces due to `Q_(1)` and `Q_(4)` on q is resolved into components.
`F_(2)sinthetaandF_(4)sintheta` i.e `F_(1)sin45^(@)andF_(2)sin45^(@)` are equal and opposite. So they get cancel.
Total force acting on q is due to `Q_(3)` (i.e `F_(3))`
`F_(4)costheta,F_(2)costhetaF=F_(3)+F_(2)costheta+F_(4)costheta`
Total force `F=k.(qQ)/(R^(2))+k.(qQ)/(R^(2)).cos45^(@)+(kqQ)/(R^(2))cos45^(@)`
`=(kqQ)/(R^(2))[1+(2)/(sqrt2)]`
Total F `=(kqQ)/(R^(2))[1+sqrt2]hati`
`=(1)/(4piepsi_(0))(qQ)/(R^(2))[1+sqrt2]hati" "[:.k=(1)/(4piepsi_(0))]`