five charges of equal amount q are placed at five corners of a regular hexagon of side 10 cm. what will be the value of the charge placed at 6th corner of the hexagon such that the electric field at the centre of hexagon is zero ?
Answers
Answer:
Hope it will help you
Explanation:
If there were 6 charges at each corner then the net electric field at the centre would be zero since it is a regular hexagon . the electric field due to 5 charges will be equal and opposite to the net electric field due to sixth charge so as to make the net electric field at the centre to be zero . Hence electric field at the centre due to 5 charges is will be equal to electric field at the centre due to sixth charge
F = K q2/a2 where K is 1/4π€
Answer:
Let O be the centre of the hexagon
It follows that the point O, when joined to the ends of a side of the hexagon forms an equilateral triangle.
∴AO=BO=CO=DO=EO=FO=10cm=0.1 m
Since at each comer of the hexagon, a charge of 5pc i.e., 5×10−6C is placed, total electric potential at point o due to the charges at the six comers,
V=6×(4π∈01×rq)
=6×9×109×0.15×10−6
=2.7×106V.