Question

six point charges are arranged at the vertices of a regular hexagon of side 1 cm. calculate the net electric field

Answer 1

Step-by-step explanation:

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.

Answer 2

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