Six point charges are kept 60 apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the center of the circle is . ( 0 is permittivity of free space) [figure]
- - ( 5Q8 0 R2 )( i - 3 j)
- Q4 0 R2( 3 i - j)
- - Q4 0 R2( 3 i - j)
- - 5Q8 0 R2( i + 3 j)
Answer
(3) - Q4 0 R2( 3 i - j)
Solution
If all six charges were +Q, the fields at the centre would cancel in pairs (opposite charges are diametrically opposite): E = 0
The actual arrangement = six +Q charges + a charge -2Q at the position of -Q (at 150 from the x-axis).
Net field = field of -2Q alone.
Field of a negative charge points towards the charge, i.e. along 150 :
n = 150 i + 150 j = - 32 i + 12 j
E = 2Q4 0 R2 (- 32 i + 12 j )
E = - Q4 0 R2( 3 i - j)
Answer: (3)