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Find the Laplace equation value of the following potential field V = r cos θ + φ
- a)3
- b)2
- c)1
- d)0
Correct answer is option 'D'. Can you explain this answer?
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Find the Laplace equation value of the following potential field V = r...
Answer: d
Explanation: (Del)2 (r cos θ + φ) = (2 cosθ/r) – (2 cosθ/r) + 0
= 0, this satisfies Laplace equation. This value is 0.
Explanation: (Del)2 (r cos θ + φ) = (2 cosθ/r) – (2 cosθ/r) + 0
= 0, this satisfies Laplace equation. This value is 0.
Most Upvoted Answer
Find the Laplace equation value of the following potential field V = r...
To find the Laplace equation value of the potential field V = r cosθ, we need to calculate the Laplacian of V, which is given by:
∇²V = (1/r) ∂/∂r (r ∂V/∂r) + (1/r²) ∂²V/∂θ²
First, let's calculate the partial derivatives of V with respect to r and θ:
∂V/∂r = cosθ
∂²V/∂θ² = -cosθ
Next, we substitute these derivatives into the Laplacian equation:
∇²V = (1/r) ∂/∂r (r cosθ) + (1/r²) (-cosθ)
= (1/r) (cosθ + r(-sinθ)) + (-cosθ)/r²
= (1/r) cosθ - sinθ + (-cosθ)/r²
Simplifying further, we have:
∇²V = (1/r) cosθ - sinθ - (cosθ)/r²
Therefore, the Laplace equation value of the potential field V = r cosθ is:
∇²V = (1/r) cosθ - sinθ - (cosθ)/r²
∇²V = (1/r) ∂/∂r (r ∂V/∂r) + (1/r²) ∂²V/∂θ²
First, let's calculate the partial derivatives of V with respect to r and θ:
∂V/∂r = cosθ
∂²V/∂θ² = -cosθ
Next, we substitute these derivatives into the Laplacian equation:
∇²V = (1/r) ∂/∂r (r cosθ) + (1/r²) (-cosθ)
= (1/r) (cosθ + r(-sinθ)) + (-cosθ)/r²
= (1/r) cosθ - sinθ + (-cosθ)/r²
Simplifying further, we have:
∇²V = (1/r) cosθ - sinθ - (cosθ)/r²
Therefore, the Laplace equation value of the potential field V = r cosθ is:
∇²V = (1/r) cosθ - sinθ - (cosθ)/r²
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Find the Laplace equation value of the following potential field V = r cos θ + φa)3b)2c)1d)0Correct answer is option 'D'. Can you explain this answer?
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