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Question

# Three charges of equal magnitude q is placed at the vertices of an equilateral triangle of side l. The force on a charge Q placed at the centroid of the triangle is

A
3Qq4πε0l2
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B
2Qq4πε0l2
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C
Qq2πε0l2
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D
zero
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Solution

## The correct option is C zeroAs shown in figure draw AD⊥BC∴AD=ABcos30o=l√32Distance AO of the centroid O from A=23AD=2l3√32=1√3∴ Force on Q at O due to charge q1=q at A→F1=14πε0Qq(l√3)2=3Qq4πε0l2,along AOSimilarly, force on O due to charge q2=q at B→F2=3Qq4πε0l2, along BOand force on Q due to charge q3=q at C→F3=3Qq4πε0l2 along COAngle between forces F2 and F3=120oBy parallelogram law, resultant of →F2 and →F3 3Qq4πε0l2 along OA∴ force on Q=3Qq4πε0l2−3Qq4πε0l2=0

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