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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????????
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
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
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