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Ankita Choudhury Grade: 12
        The potential energy of a particle in a certain  field has the form U=a/r to the power 2-  b/r,where a  and b are '+'ve constants and r is distance from centre of the field.find r corresponding to equilibrium position of particle????????    
7 years ago

Answers : (1)

AskIITians Expert Hari Shankar IITD
17 Points
										

Hi,


At equilibrium, the particle will try to have minimum potential energy.


To find the minimum of any function f with respect to x, we need to put df/dx = 0.


Here, we need to find the minimum of the potential energy function U = a/r2 - b/r.


So, we have to differentail U with respect to r,


and then put dU/dr = 0 to find out the position where U is minimum.


U = ar-2 -br-1


dU/dr = -2ar-3 - (-br-2) = (-1/r2)(2a/r - b)


Putting dU/dr = 0,


(-1/r2)(2a/r - b) = 0


or, 2 a/r - b = 0, because 1/r2 = 0 will give r = infinity, which is anyway true


Hence, r = b/2a is the equilibrium distance

7 years ago
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