1.

Three charges are placed on the circumference of a circle of radius d as shown in the figure. Find the electric field along x-axis at the centre of the circle : Electric field due to -4q vecE_(1) =(4kq)/d^(2) electric field due to +2q and -2q vecE_(23) = (4kq)/d^(2)

Answer»

`q/(4piepsilon_(0)d^(2))`
`(qsqrt(3))/(4piepsilon_(0)d^(2))`
`(qsqrt(3))/(piepsilon_(0)d^(2))`
`(qsqrt(3))/(2piepsilon_(0)d^(2))`

Solution :Y-components of `vecE_(1)`and `vecE_(23)`are equal in magnitude and opposite in DIRECTION hence RESULTANT ELECTRIC FIELD can be obtained by addition of jc-components only. `therefore` Resultant electric field
`E==(4kq)/d^(2)cos30^(@) + (4kq)/d^(2) cos(-30^(@))`
`=q/(piepsilon_(0)d^(2)) xx (2 xx SQRT(3))/2 =(qsqrt(3))/(piepsilon_(0)d^(2))`


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