Two point charges +q and -2qare placed at the vertices B and C of an equilateral triangle ABC of side 'a' as shown in Obtain the expression for
The magnitude ,
(b)
The direction of the resultant electric field at the vertex A due to these two charges.
(c)

Explanation

As shown in electric field at vertex A due to charge +q placed at vertex B is
` " " E_B =(1)/( 4 pi in _0) . (q)/( a^(2)) ` along BA
and electrical field at vertex A due to charges -2q placed at vertex C is
` " " E_C =(1)/( 4 pi in _0) (2q)/( a^(2)) ` along AC
Obviously ` oversetto (E_B) and oversetto (E_C) ` are inclined at an angle ` theta = 120^(@) ` from the another . Hence
Magnitude of net electric field at A
` " " E= sqrt( E_B^(2) + E_C^(2)+2E_BE_C cos 120 ^(@) )= sqrt(E_B^(2) +E_C^(2) -E_B E_C ) `
` rArr " " E= ( 1)/( 4 pi in _0) .(sqrt3q)/( a^(2)) `
(b) If the net electric field `oversetto E ` is directed at angle `beta ` from the direction AC, then
` tan beta = ( E_Bsin 120^(@) )/(E_C+ E_B cos 120 ^(@)) =(E_B.(sqrt3//2))/(E_C-(E_B//2))=(E_B.(sqrt3//2))/(2E_B-(E_B//2))`
`rArr " " tan beta = (1)/(sqrt3) or beta= tan ^(-1) ((1)/(sqrt3)) =30^(@) `