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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

Text Solution

Verified by Experts

The correct Answer is:
C
The force experienced by A due to B is
`F_1=(1)/(4piepsi_0)("QQ")/(a^2) " along " bar(AB)` (attractive)
The force experienced by A due to C is
`F_2=(1)/(4piepsi_0)(Q^2)/(a^2) " along " bar(CA)` produced (repulsive)
Now the force `vecF_1` is having a component `F_1cos30^@` along and the force `vecF_2` is having a component `F_2cos30^@ " along " bar(AE). bar(AE) " and " bar(AF)` are both normal to `bar(BC)` but they are mutually opposite to each other. And as `|vecF_1|=|vecF_2|`, so the forces along `bar(AE) " and " bar(AF)` will both cancel each other. And so the force experienced by the charge at A in the direction normal to `bar(BC)` is zero .
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