In cylindrical coordinates, the Laplace equation holds the following expression :
\(\nabla^{2}V=\dfrac{1}{\rho}\dfrac{\partial}{\partial\rho}\left(\rho\dfrac{\partial V}{\partial\rho}\right)+\dfrac{1}{\rho^{2}}\dfrac{\partial^{2}V}{\partial\phi^{2}}+\dfrac{\partial^{2}V}{\partial z^{2}}=0\)
Laplace's equation in cylindrical coordinates must contain all three coordinates, and only option 3 does.
\(\nabla^{2}V=\dfrac{1}{\rho}\dfrac{\partial}{\partial\rho}\left(\rho\dfrac{\partial V}{\partial\rho}\right)+\dfrac{1}{\rho^{2}}\dfrac{\partial^{2}V}{\partial\phi^{2}}+\dfrac{\partial^{2}V}{\partial z^{2}}=0\)
Two features identify the correct form, and each eliminates a distractor.
The factor \(\rho\) inside the radial derivative. The Laplacian is the divergence of the gradient, and in cylindrical coordinates the divergence carries a \(1/\rho\) with a \(\rho\) inside:
\(\nabla\cdot\vec{A}=\dfrac{1}{\rho}\dfrac{\partial\left(\rho A_{\rho}\right)}{\partial\rho}+\cdots\)
That \(\rho\) is a metric factor: a shell at radius \(\rho\) has area proportional to \(\rho\), so flux through it grows with radius even when the field does not. Option 1 omits it and so is not a Laplacian at all.
The factor \(1/\rho^{2}\) before the angular term. The angle \(\phi\) is dimensionless, so \(\partial^{2}V/\partial\phi^{2}\) has units of volts, not volts per square metre. Dividing by \(\rho^{2}\) restores the dimensions — and physically it says that at large radius a given angular change corresponds to a greater arc length \(\rho\,d\phi\), so the same angular variation produces a smaller field.
| Option | Fault |
|---|---|
| 1 | Missing the ρ inside the radial derivative, and no z term |
| 2 | Radial and z terms correct, but the φ term is absent |
| 3 | ✓ All three terms with the right metric factors |
| 4 | Only the z term — this is the one-dimensional Laplace equation |
A quick way to remember the whole family. Each coordinate contributes a second derivative divided by the square of the length element in that direction: \(d\rho\), \(\rho\,d\phi\) and \(dz\). Where the length element depends on position, an extra factor appears — which is precisely the \(\rho\) and the \(1/\rho^{2}\). In Cartesian coordinates all three length elements are constant, which is why that form is the simple sum of three second derivatives.
Where it is used : the potential in a coaxial cable, around a charged wire, and inside a cylindrical waveguide all follow from this equation with the appropriate boundary conditions — the waveguide case producing the Bessel functions that give the guide its mode structure.
Hence, the correct expression is option 3.
Similar Questions
A scalar function V is given by V = 2xyz2. The gradient of V is given by:
Assertion (A) : Curl of any vector is a vector. It gives normal vector which is perpendicular to both the plane and the parent vector.
Reason (R) : The value of curl of H can be found by the expression :
(Curl H)Normal = \(\lim_{\Delta S\to0}\dfrac{\oint \vec{H}\cdot d\vec{l}}{\Delta S}\)
where $\Delta S$ is the planar area and $\overline{\text{dl}}$ is the line element.
Select your answer using the codes given below.
Match the following :
List - I List - II (a) Curl operator (i) Gradient (b) Del operator (ii) Volume to surface conversion (c) Divergence tdeorem (iii) Surface to line conversion (d) Stokes tdeorem (iv) Rotation Codes :
\(\nabla\times\nabla\times\vec{A}\) is equal to :
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