The potential energy of a certain field has the form U=a/r^2-b/r,where a and b are positive constants, r is the distance from the centre of the field. Find the value of R corresponding to the equilibrium position of the particle; examine whether this position is stable.
Asked by devanshu_jain | 19th Oct, 2010, 07:26: PM
The equilibrium position is one with minimum potential energy, hence,
dU/dr = 0
-2ar-3 + br-2 = 0
r = 2a/b
r = R = 2a/b
To investigate the stability of this position, we need to find the curve is concave or convex at R.
d2U/dr2 = 6ar-4 -2br-3
at r = R
d2U/dr2 = 6a(2a/b)-4 -2b(2a/b)-3
= 6b4/(16a3) -2b4/(8a3)
= b4/(8a3) > 0
Hence the this position is concave and stable.
Answered by | 19th Oct, 2010, 10:12: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number