A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is [figure]
- E along KO
- 3E along OK
- 3E along KO
- E along OK
Answer
(4) E along OK
Solution
By symmetry, the field at the centre due to the whole uniformly charged ring is zero.
EAKB + EACDB = 0 EACDB = - EAKB
The positive charge on AKB gives a field E at O directed away from it, i.e. along KO.
The field due to ACDB has the same magnitude E and is directed opposite, i.e. along OK.
Answer: (4)