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Equal charges q are placed at the vertic...

Equal charges `q` are placed at the vertices `A` and `B` of an equilatral triangle `ABC` of side `a`. The magnitude of electric field at the point `C` is

A

E along AO

B

2E along AO

C

E along BO

D

E along CO

Text Solution

Verified by Experts

The correct Answer is:
A
Resolve E along CO and BO into two perpendicular components

The sine components cancel each other
The cosine components add up along OA to give `2E cos 60^@`
`:.` Resultant field along `AO = 2E - 2E cos 60^@`
`=2E - E = E`
`:.` Resultant field is E along AO .
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