A thin conducting ring orf radius `R` is given a chareg `+Q`, Fig. 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 part `ACDB` of the ring is

Options

A. `3 E` along `KO`
B. `E` along `OK`
C. `E` along `KO`
D. `3E` along `OK`

Answer

B

Explanation

Charge on part `AKB = Q//4`, Fig
`:. E` at O due to `AKB` is due to charge `Q//4`.
In the part `ACDB`, intensity at `O` due to charges on `AC` and `DB` cancels out. Net intensity at `O` is due to charge `Q//4` on part `CD`. Its value must be `E` along OK.