Purcell Problem 2.11

Physics · electromagnetism

Problem

The Laplacian We have now met two scalar functions related to the electric field, the potential function φ\varphi (see Eq. (2.16)) and the divergence, div E. In Cartesian coordinates the relationships are expressed as follows: E=−grad φ=−(x^∂φ∂x+y^∂φ∂y+z^∂φ∂z),(2.66)E = -\text{grad } \varphi = -\left( \hat{x} \frac{\partial \varphi}{\partial x} + \hat{y} \frac{\partial \varphi}{\partial y} + \hat{z} \frac{\partial \varphi}{\partial z} \right), \quad (2.66) div E=∂Ex∂x+∂Ey∂y+∂Ez∂z.(2.67)\text{div } E = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}. \quad (2.67) Equation (2.66) shows that the x component of E is Ex=−∂φ∂xE_x = -\frac{\partial \varphi}{\partial x}. Substituting this and the corresponding expressions for EyE_y and EzE_z into Eq. (2.67), we get a relation between div E and φ\varphi: div E=−div grad φ=−(∂2φ∂x2+∂2φ∂y2+∂2φ∂z2).(2.68)\text{div } E = -\text{div grad } \varphi = -\left( \frac{\partial^2 \varphi}{\partial x^2} + \frac{\partial^2 \varphi}{\partial y^2} + \frac{\partial^2 \varphi}{\partial z^2} \right). \quad (2.68) The operation on φ\varphi that is indicated by Eq. (2.68), except for the minus sign, we could call "div grad," or "taking the divergence of the gradient of ...." The symbol used to represent this operation is ∇2\nabla^2, called the Laplacian operator, or just the Laplacian. The expression ∂2∂x2+∂2∂y2+∂2∂z2(2.69)\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \quad (2.69) is the prescription for the Laplacian in Cartesian coordinates. So we have div E=−∇2φ(2.70)\text{div } E = -\nabla^2 \varphi \quad (2.70) The notation ∇2\nabla^2 is explained as follows. With the vector operator ∇\nabla given in Eq. (2.60), its square equals ∇⋅∇=∂2∂x2+∂2∂y2+∂2∂z2,(2.71)\nabla \cdot \nabla = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}, \quad (2.71) the same as the Laplacian in Cartesian coordinates. So the Laplacian is often called "del squared," and we say "del squared φ\varphi," meaning "div grad φ\varphi." Warning: In other coordinate systems, spherical coordinates, for instance, the explicit forms of the gradient operator and the Laplacian operator are not so simply related. This is evident in the list of formulas at the beginning of Appendix F. It is well to remember that the fundamental definition of the Laplacian operation is "divergence of the gradient of." We can now express directly a local relation between the charge density at some point and the potential function in that immediate neighborhood. Combining Eq. (2.70) with Gauss's law in differential form, div E = ρ/ϵ0\rho/\epsilon_0, we have ∇2φ=−ρϵ0(2.72)\nabla^2 \varphi = -\frac{\rho}{\epsilon_0} \quad (2.72) Equation (2.72), sometimes called Poisson's equation, relates the charge density to the second derivatives of the potential. Written out in Cartesian coordinates it is ∂2φ∂x2+∂2φ∂y2+∂2φ∂z2=−ρϵ0.(2.73)\frac{\partial^2 \varphi}{\partial x^2} + \frac{\partial^2 \varphi}{\partial y^2} + \frac{\partial^2 \varphi}{\partial z^2} = -\frac{\rho}{\epsilon_0}. \quad (2.73) One may regard this as the differential expression of the relationship expressed by the integral in Eq. (2.18), which tells us how to find the potential at a point by summing the contributions of all sources near and far. Example (Poisson's equation for a sphere) Let's verify that Eq. (2.72) holds for the potential due to a sphere with radius R and uniform charge density ρ\rho. This potential was derived in the second example in Section 2.2. Spherical coordinates are the best choice here, so we will invoke the expression for the Laplacian in spherical coordinates, given in Eq. (F.3) in Appendix F. Since the potential depends only on r, we have ∇2φ=(1/r2)∂∂r(r2∂φ∂r)\nabla^2 \varphi = (1/r^2) \frac{\partial}{\partial r}(r^2 \frac{\partial \varphi}{\partial r}). The potential outside the sphere is φ=ρR33ϵ0r\varphi = \frac{\rho R^3}{3\epsilon_0 r}. All that matters here is the fact that φ\varphi is proportional to 1/r, because this makes ∂φ∂r\frac{\partial \varphi}{\partial r} proportional to 1/r², from which we immediately see that ∇2φ=0\nabla^2 \varphi = 0. This agrees with Eq. (2.72), because ρ=0\rho = 0 outside the sphere. Inside the sphere, we have φ=ρR22ϵ0−ρr26ϵ0\varphi = \frac{\rho R^2}{2\epsilon_0} - \frac{\rho r^2}{6\epsilon_0}. The constant term vanishes when we take the derivative, so we have ∇2φ=1r2∂∂r(r2∂φ∂r)=1r2∂∂r(−ρr3ϵ0)=1r2⋅(−ρr23ϵ0)=−ρϵ0,(2.74)\nabla^2 \varphi = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \varphi}{\partial r}\right) = \frac{1}{r^2} \frac{\partial}{\partial r}\left(-\frac{\rho r}{3\epsilon_0}\right) = \frac{1}{r^2} \cdot \left(-\frac{\rho r^2}{3\epsilon_0}\right) = -\frac{\rho}{\epsilon_0}, \quad (2.74) as desired. 86 The electric potential

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