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# 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???????? AskIITians Expert Hari Shankar IITD
17 Points
11 years ago

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