Question

Two similar and identical metal spheres $A$ and $B$ repel each other with a force of $2\times {10}^{-5}N$. AThird identical uncharged sphere $C$ is touched with $A$ and then placed at the mid-point of $AB$. What is the net electric force on $C$?

• A. $5\times {10}^{-5}N$
• B. $2\times {10}^{-5}N$
• C. $3\times {10}^{-5}N$
• D. $4\times {10}^{-5}N$

Answer 1

The correct option is B $2\times {10}^{-5}N$

$LetthechargeonAandBbeQeachLetthedistancebetweenAandBis2RAccordingtoColoum{b}^{\prime }slaw,ForcebetweenAandBisF=\frac{K{Q}^2}{{\left(2R\right)}^2}TheforcebetweenAandBis2\times {10}^{-5}Nor2\times {10}^{-5}N=\frac{K{Q}^2}{4R^2}=F_{1}WhenanunchargedsimillarsphereCistouchedtoA,itschargewillbeequallydistributedbetweenAandC.ie;\frac{Q}{2}CisplacedatthemidpointofAandB,ie;atdistanceRfrombothside.NowforcebetweenAandCis,F_2=\frac{K\frac{Q}{2}\frac{Q}{2}}{R^2}=\frac{K\frac{Q^2}{4}}{R^2}=\frac{K{Q}^2}{4R^2}ForcebetweenBandCis{F}_3=\frac{K\frac{Q}{2}\times Q}{R^2}=\frac{K{Q}^2}{2R^2}Wecanseethat,F_3>F_{2}Therefore,ResultantforceonCis{F}_3-F_2=\frac{K{Q}^2}{2R^2}(1-\frac{1}{2})=\frac{K{Q}^2}{2R^2}\times \frac{1}{2}=\frac{K{Q}^2}{4R^2}whichisequalto=2\times {10}^{-5}N$