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Chapter 6: Problem 46

By appropriate solution of Laplace's and Poisson's equations, determine the absolute potential at the center of a sphere of radius \(a\), containing uniform volume charge of density \(\rho_{0}\). Assume permittivity \(\epsilon_{0}\) everywhere. Hint: What must be true about the potential and the electric field at \(r=0\) and at \(r=a\) ?

Short Answer

Expert verified
Answer: The absolute potential at the center of the uniformly charged sphere is 0.

Step by step solution

01

Understand the given information and boundary conditions

We have a uniformly charged sphere with radius \(a\) and charge density \(\rho_0\). We need to find the absolute potential, which is given by the Laplace's equation or Poisson's equation: Laplace's equation: \(\nabla^2 V = 0\) Poisson's equation: \(\nabla^2 V = -\frac{\rho}{\epsilon_0}\) Since we have a charge density, we use Poisson's equation. For a spherical system, the equation is given in spherical coordinates as: \(\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial V}{\partial r}\right)=-\frac{\rho_0}{\epsilon_0}\) We need to solve this equation subject to the boundary conditions at \(r=0\) and \(r=a\).
02

Solve Poisson's equation in spherical coordinates

Integrate the given Poisson's equation twice with respect to \(r\) to find the potential: \(\frac{\partial}{\partial r}\left(r^2\frac{\partial V}{\partial r}\right)=-\frac{\rho_0}{\epsilon_0}r^2 + C_1\) \(\frac{\partial V}{\partial r}= -\frac{\rho_0}{3\epsilon_0}r^3 + C_1r + C_2\) Now, integrate once more: \(V(r)= -\frac{\rho_0}{12\epsilon_0}r^4 + \frac{C_1}{2}r^2 + C_2r + C_3\)
03

Apply the boundary conditions

At \(r=0\), we must have a finite potential and electric field, which implies that the potential must follow: \(\lim_{r \to 0}V(r)=\lim_{r \to 0}\frac{\partial V}{\partial r}=0\) From the potential equation, it implies that \(C_2=0\). At \(r=a\), the potential must be continuous, which means that: \(V(a)= -\frac{\rho_0}{12\epsilon_0}a^4 + \frac{C_1}{2}a^2 + C_3\)
04

Solve for the constants \(C_1\) and \(C_3\)

To find the potential at any point within the sphere, we need to determine the constants \(C_1\) and \(C_3\). We have: \(-\frac{\rho_0}{12\epsilon_0}a^4 + \frac{C_1}{2}a^2 + C_3\) For a uniform charge distribution, the potential at the surface of the sphere is: \(V(a) = \frac{1}{4\pi\epsilon_0}\frac{1}{a}\frac{a^3}{3}\rho_0 = \frac{\rho_0 a^2}{12\epsilon_0}\) Now we can solve the constants: \(C_1 = \frac{3}{2}\rho_0 a^2\) To find \(C_3\), we plug \(C_1\) back into the equation for the potential at the surface: \(-\frac{\rho_0}{12\epsilon_0}a^4 + \frac{3}{2}\rho_0 a^4 + C_3 = \frac{\rho_0 a^2}{12\epsilon_0}\) Solving for \(C_3\) yields: \(C_3 = 0\)
05

Calculate the potential at the center of the sphere

Now that we have found the constants \(C_1\) and \(C_3\), we can find the potential at the center of the sphere (\(r=0\)): \(V(0)=-\frac{\rho_0}{12\epsilon_0}(0)^4+\frac{3}{2}\rho_0(0)^2+0=0\) So, the absolute potential at the center of a uniformly charged sphere is 0.

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Most popular questions from this chapter

(a) Determine the capacitance of an isolated conducting sphere of radius \(a\) in free space (consider an outer conductor existing at \(r \rightarrow \infty\) ). \((b)\) The sphere is to be covered with a dielectric layer of thickness \(d\) and dielectric contant \(\epsilon_{r}\). If \(\epsilon_{r}=3\), find \(d\) in terms of \(a\) such that the capacitance is twice that of \(\operatorname{part}(a) .\)

Construct a curvilinear-square map of the potential field between two parallel circular cylinders, one of \(4 \mathrm{~cm}\) radius inside another of \(8 \mathrm{~cm}\) radius. The two axes are displaced by \(2.5 \mathrm{~cm}\). These dimensions are suitable for the drawing. As a check on the accuracy, compute the capacitance per meter from the sketch and from the exact expression: $$C=\frac{2 \pi \epsilon}{\cosh ^{-1}\left[\left(a^{2}+b^{2}-D^{2}\right) /(2 a b)\right]}$$ where \(a\) and \(b\) are the conductor radii and \(D\) is the axis separation.

A coaxial cable has conductor dimensions of \(a=1.0 \mathrm{~mm}\) and \(b=2.7 \mathrm{~mm}\). The inner conductor is supported by dielectric spacers \(\left(\epsilon_{r}=5\right)\) in the form of washers with a hole radius of \(1 \mathrm{~mm}\) and an outer radius of \(2.7 \mathrm{~mm}\), and with a thickness of \(3.0 \mathrm{~mm}\). The spacers are located every \(2 \mathrm{~cm}\) down the cable. ( \(a\) ) By what factor do the spacers increase the capacitance per unit length? \((b)\) If \(100 \mathrm{~V}\) is maintained across the cable, find \(\mathbf{E}\) at all points.

The hemisphere \(0

Two conducting spherical shells have radii \(a=3 \mathrm{~cm}\) and \(b=6 \mathrm{~cm} .\) The interior is a perfect dielectric for which \(\epsilon_{r}=8\). (a) Find \(C .(b)\) A portion of the dielectric is now removed so that \(\epsilon_{r}=1.0,0<\phi<\pi / 2\), and \(\epsilon_{r}=8\), \(\pi / 2<\phi<2 \pi\). Again find \(C\).
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