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The potential energy of a particle in a certain fieldhas the form U=a/r^2-b/r,where a and b are positive constants, r is the distance from the center of the field. Then |
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Answer» At `r=(2a)/B`,particle is in STEADY equilibriumn. For `(dU)/(dr)=0, b/r^2=(2a)/r^3 rArr r=(2a)/b` `(d^2U)/(dr^2)=(+6a)/r^4-(2B)/r^3=2/r^3((3a)/r-b)` At `r=(2a)/b, (d^2U)/(dr^2)=2/r^3 ((3axxb)/(2a)-b)=2/r^3 b/2=b/r^3 GT 0` i.e., U is minimum So, it is a positionof stable (steady) equilibrium. `F=-(dU)/(dr)=(2a)/r^3-b/r^2` For maximum force , `(DF)/(dr)=(-d^2U)/(dr^2)=0` `rArr (-2)/r^3 ((3a)/r-b)=0 rArr r=(3a)/b` `F=(2a)/((3a)/b)^3-b/((3a)/b)^2=(2ab^3)/(27a^3)-b^3/(9a^2)=-b^3/(27a^2)` |
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