Question

Six point charges are placed at the vertices of a Hexagon of side 1m are-q -q -q +q +q +q respectively net electric field at the centre of the hexagon

Answer 1

magnitude of electric field due to each charge at centre of hexagon, E = kq²/(1)² = kq²

Let ABCDEF is regular hexagon and O is the centre of it as shown diagram,

electric field due to D on O is along electric field due to A on O

so, net electric field due to A and D on O,$E_1$  = E + E = 2E along OA [ see figure ]

similarly, electric field due to E and B,$E_2$ = E + E = 2E along OB [ see figure ]

electric field due to F and C, $E_3$ = E + E = 2E along OC [ see figure ]

now, resultant of $E_1$ and $E_3$ = 2E along OB [ because angle between them is 120° ]

now, net electric field an O = resultant of $E_1$ and $E_3$ + electric field due to E and B on O

= 2E + 2E = 4E along OB

= 4kq² [ as E = kq² ]

hence, net electric field at O = 4kq² along OB.

Answer 2

Answer:

magnitude of electric field due to each charge at centre of hexagon, E = kq²/(1)² = kq²

Let ABCDEF is regular hexagon and O is the centre of it as shown diagram,

electric field due to D on O is along electric field due to A on O

so, net electric field due to A and D on O,  = E + E = 2E along OA [ see figure ]

similarly, electric field due to E and B, = E + E = 2E along OB [ see figure ]

electric field due to F and C,  = E + E = 2E along OC [ see figure ]

now, resultant of  and  = 2E along OB [ because angle between them is 120° ]

now, net electric field an O = resultant of  and  + electric field due to E and B on O

= 2E + 2E = 4E along OB

= 4kq² [ as E = kq² ]

hence, net electric field at O = 4kq² along OB.

Explanation: