Four point charges each of value +q are placed on five vertices of regular Pentagon of side L.what is the magnitude of the force on a point charge of value -q placed at the centre of the hexagon
Answers
Given info : Four point charges each of value +q are placed on five vertices of regular Pentagon of side L. Another point charge -q is placed at centre of pentagon.
To find : force on the charge placed at centre of the pentagon.
solution : all four charges placed on the vertices of pentagon exerted same force on the centre.
and force exerted by each charge placed at vertex of pentagon is given by, F = kq²/L²
There are two forces (A and B) made 72° with vertical lines as shown in figure.
so, resultant of these forces , F' = 2Fcos72°
The other two force made 72° with each other. So resultant force of these must be along middle line (vertically downward) of these.
and resultant of these forces, F" = 2Fcos36°
Now net force acting on the charge -q , Fnet = 2Fcos72° - 2Fcos36°
= 2F[cos72° -cos36°]
= 2F[2sin54° sin18° ]
= 4Fsin54° sin18°
= 4 F × 0.25
= F
= Kq²/L²
Therefore the net force acting on the charge -q placed at the centre of pentagon is kq²/L²
