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To determine whether the potential ( V = 25 sin theta ) satisfies Laplace’s equation, we need to check whether it satisfies the equation ( nabla^2 V = 0 ).
In spherical coordinates, Laplace’s equation is given by:
[
nabla^2 V = frac{1}{r^2} frac{partial}{partial r} left( r^2 frac{partial V}{partial r} right) + frac{1}{r^2 sin theta} frac{partial}{partial theta} left( sin theta frac{partial V}{partial theta} right) + frac{1}{r^2 sin^2 theta} frac{partial^2 V}{partial phi^2} = 0
]
Since ( V ) only depends on ( theta ) and not on ( r ) or ( phi ), we can ignore the terms involving ( r ) and ( phi ). We only need to calculate the angular part:
1. Calculate ( frac{partial V}{partial theta} ):
[
frac{partial V}{partial theta} = 25 cos theta
]
2. Calculate ( frac{partial}{partial theta} left( sin theta frac
Answer: 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.