Electrostatic Potential And Capacitance Ques 1
1. The potential energy of particle in a force field is $U=\frac{A}{r^2}-\frac{B}{r}$, where $A$ and $B$ are positive constants and $r$ is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is:
$[2012]$
(a) $B / 2 A$
(b) $2 A / B$
(c) $A / B$
(d) $B / A$
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Answer:
Correct Answer: 1.(b)
Solution: (b) for equilibrium
$ \frac{d U}{d r}=0 \quad \Rightarrow \frac{-2 A}{r^3}+\frac{B}{r^2}=0 \quad r=\frac{2 A}{B} $
for stable equilibrium
$\frac{d^2 U}{d r^2}$ should be positive for the value of $r$.
Here $\frac{d^2 U}{d r^2}=\frac{6 A}{r^4}-\frac{2 B}{r^3}$ is $+ve $ value for
$ r=\frac{2 A}{B} $
After displacing a charged particle from its equilibrium position, if it return back then it is said to be in stable equilibrium. If $\mathrm{U}$ is the potential energy then in case of stable equilibrium $\frac{d_2 U}{d x^2}$ is positive i.e., $\mathrm{U}$ is minimum.
BETA
