Question

Three point charges Q, 2Q and 8Q are placed on a straight line 9 cm long. Charges are placed in
such a way that the system has minimum potential energy. Then

• A. 2Q and 8Q must be at the ends and Q at a distance of 3 cm from the 8Q.
• B. 2Q and 8Q must be at the ends and Q at a distance of 6 cm from the 8Q.
• C. Electric field at the position of Q is zero.
• D. Electric field at the position of Q is $\frac{Q}{4\pi {\epsilon }_0}$

Answer 1

Answer: (B), (C)

The correct options are
B 2Q and 8Q must be at the ends and Q at a distance of 6 cm from the 8Q.
C Electric field at the position of Q is zero.

$OnthestraightlinethechargesofgreatermagnitudeshouldbepresentattheoppositeendsnowletthepositionofchargeQbexcmdistancefrom8Qchargethandistanceofitfrom2Qbecomes9-xcmTheexpressionoftotalpotentialenergyUbecomes:U=\frac{K8Q^2}{x}+\frac{K2Q^2}{9-x},ForminimumpotentialenergydU/dx=0\frac{dU}{dx}=\frac{-4}{x^2}+\frac{1}{{(9-x)}^2}=0,x=6,18x=18isnotpermissibleinthisquestionThereforex=6cm,distanceofQfrom8Qis6cmsooptionBTotalelectricfieldEatx=6cmdueto8Qand2Qpointchargesisgivenas:E=\frac{8K{Q}^2}{6^2}-\frac{2K{Q}^2}{3^2}=0SOoptionC$