In the rectangle shown below the two corners have charges q1=−5μC and q2=+2.0μC The work done in moving a charge +3.0μC From B to A is (take 1/4πε0=1010 N−m2/C2 )
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a
2.8 J
b
2.5 J
c
3.2 J
d
3.5 J
answer is A.
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Detailed Solution
Work doneVA=1010−5×10−615×10−2+2×10−65×10−2=115×106 volt and VB=10102×10−615×10−2−5×10−65×10−2=−1315×106 volt ∴WBA= QVA- VB=3×10−6115×106−−1315×106=2.8 J