Poll Results
No votes. Be the first one to vote.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
To determine if the given potential ( V = 25 sin theta ) satisfies Laplace’s equation in free space, we need to utilize the form of Laplace’s equation in spherical coordinates, as the potential is given in terms of (theta), which is a spherical coordinate.
Laplace’s equation in spherical coordinates (assuming azimuthal symmetry, which seems to be implied here as the potential is a function of (theta) alone) is:
[
nabla^2V = frac{1}{r^2}frac{partial}{partial r}left( r^2frac{partial V}{partial r} right) + frac{1}{r^2sintheta}frac{partial}{partial theta}left( sinthetafrac{partial V}{partial theta} right) + frac{1}{r^2sin^2theta}frac{partial^2 V}{partial phi^2} = 0
]
Given ( V = 25 sin theta ), this does not depend on ( r ) or ( phi ), so the first and third terms in the Laplace equation vanish. We’re only left with the second term:
[
frac{1}{r^2sintheta}frac{partial}{partial theta}left( sinthetafrac{
a
Explanation: (Del)2V = 0
(Del)2V = (Del)2
(25 sin θ), which is not equal to zero. Thus the field does not satisfy
Laplace equation.