Five charges each `q` are placed at five corners regular pentagon. Distance from corner to the centre of pontagon is `r`. Then `(K=(1)/(4pi epsilon_(0)))`

Assertion : Five charges +q each are placed at five vertices of a regular pentagon. A sixth xharge -Q is placed at the centre of pentagon. Then, net electrostatic force on -Q is zero. Reason : Net electrostatic potential at the centre is zero.

Five equal charges each of value q are placed at the corners of a regular pentagon of side a. The electric field at the centre of the pentagon is

Five equal charges each of value q are placed at the corners of a regular pentagon of side a. The electric field at the centre of the pentagon is

Six charges +q each are placed at the corners of a regular hexagon of each side a. The electric field at the centre of the hexagon is

Five charges , q each are placed at the corners of a regular pentagon of side. (a) (i) What will be the electric field at 0, the centre of the pentagon? (ii) What will be the electric field at 0 if the charge from one of the corners (say A) is removed? (iii). What will be the electric field at 0 if the pentagon is replaced by n-sided regular polygon with charge q at each of its corner?

Five point charges, +q each are placed at the five vertices of a regular hexagon. The distance of the center of the hexagon from any fo the vertices is a. the electric field at the center of the hexagon is