A thin conducting ring of radius `R` is given a charge `+Q`, Fig. The electric field at the center `O` of the ring due to the charge on the part `AKB` of the ring is `E`. The electric field at the center due to the charge on part `ACDB` of the ring is

Options

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

Answer

b

Explanation

According to symmetry net electric field due to part `AKB` and the part `ACDB` at its centre is zero, i.e.,

`vec(E )_(t otal)=0`
or `vec(E )_(AKB)+vec(E )_(ACDB)=0implies vec(E )_(1)+vec(E)_(2)=0`
or `vec(E )_(2)=-vec(E )_(1)`
or `vec(E )_(ACBD)= -vec(E )` (along `KO`)