Two point charges `+qand-2q` are placed at the vertices 'B' and 'C' of an equilateral triangle ABC of side 'a' as given in the figure . Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex A due to these two charges .

Explanation

(i) The magnitude
`|vecE_(AB)|=(1)/(4piepsilon_(0))(q)/(a^(2))=E`
`|vecE_(AC)|=(1)/(4piepsilon_(0))(2q)/(a^(2))=2E`

`E_("net")=sqrt((2E)^(2)+E^(2)+2xx2ExxExx(-(1)/(2)))`
`=sqrt(4E^(2)+E^(2)-2E^(2))`
`=Esqrt(3)=(1)/(4piepsilon_(0))(qsqrt(3))/(a^(2))`
(ii) The direction of resultant electric field at vertex A
`tanalpha=(E_(AS)sin120^(@))/(E_(AC)+E_(AB)cos120^(@))`
`=(Exx(sqrt(3))/(2))/(2E+Exx((-1)/(2)))=(1)/(sqrt(3))`
`alpha=30^(@)` (with side AC)