The haplace's equation in spherical polar coordinates is :
$ [\frac{\partial}{\partial r} (\frac{1}{r^2} \frac{\partial U}{\partial r}) + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} (\sin \theta \frac{\partial U}{\partial \theta}) + \frac{1}{ \sin^2 \theta} \frac{\partial^2 U}{\partial \phi^2}] \frac{1}{r^2}= 0 $
The given question asks for the Laplace's equation in spherical polar coordinates. In mathematical physics, the Laplace equation is a second-order partial differential equation often encountered in various fields such as electrostatics, fluid dynamics, and quantum mechanics. Its general form is:
\(\nabla^2 U = 0\)where \(U\) is a scalar potential function.
In spherical polar coordinates (r, \theta, \phi), the Laplacian takes the form:
\[\nabla^2 U = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial U}{\partial r}\right) + \frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial U}{\partial \theta}\right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 U}{\partial \phi^2} = 0\]Analyzing the options given:
- Option 1: This considers the division by r^2, consistent with the Laplacian in this coordinate system.
- Option 2: Divides by r, which is not consistent with the spherical polar Laplacian.
- Option 3: Also divides by r, similarly incorrect.
- Option 4: Matches our derived expression for the Laplace's equation in spherical coordinates, with division by r^2.
Conclusion: The correct answer is Option 4, which accurately represents the Laplace equation in spherical polar coordinates:
\[\left[\frac{\partial}{\partial r} \left(\frac{1}{r^2} \frac{\partial U}{\partial r}\right) + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial U}{\partial \theta}\right) + \frac{1}{\sin^2 \theta} \frac{\partial^2 U}{\partial \phi^2}\right] \frac{1}{r^2}= 0\]Important Questions from Miscellaneous
- List-I with List-II:
List-I
(Relief Materials)List-II
(Color Code)A. Shelter I. Yellow B. Clothes II. Khaki C. Food III. Red D. Medicine IV. Brown
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