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Three charges are placed at the vertices...

Three charges are placed at the vertices of an equilateral trianlge of side `a` as shown in the following figure. The force experienced by the charge placed at the vertex `A` in a direction normal to `BC` is

A

`(Q^(2))/(4pi in_(0)a^(2))`

B

`-(Q^(2))/(4pi in_(0) a^(2))`

C

zero

D

`(Q^(2))/(2pi in_(0)a^(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
`F_(R )=2F cos theta//2 `
`=2F cos 120//2=2F cos 60`
`=F= (kQ^(2))/(a^(2))= (Q^(2))/(4pi epsilon_(0)a^(2))`

However, this force is the resultant force parallel to BC.
`:.` Force perpendicular to BC is zero.
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