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Two identical charges are placed at the two corners of an equilateral triangle . The potential energy of the system is $\;\sqcup\;.$ The work done in bringing an identical charge from infinity to the third vertex is

$(a)\;\sqcup\qquad(b)\;2\sqcup\qquad(c)\;3 \sqcup\qquad(d)\;4 \sqcup$

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Answer : (b) $\;2 \sqcup$
Explanation :
$\sqcup =\large\frac{k\;q^2}{a}$
Where q is charge on particles
and a is side of triangle
$\sqcup_{f}=\large\frac{3\;k\;q^2}{a}=3 \sqcup$
$\bigtriangleup \sqcup=2 \sqcup$
$Work \;done=2 \sqcup\;.$
answered Feb 10, 2014 by yamini.v

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