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Chapter 21: Problem 69

A solid sphere of radius \(R\) carries volume charge density \(\rho=\rho_{0} e^{r / R},\) where \(\rho_{0}\) is a constant and \(r\) is the distance from the center. Find an expression for the electric field strength at the sphere's surface.

Short Answer

Expert verified
The electric field strength at the sphere's surface is given by \(E = \frac{\rho_{0}\,R}{\epsilon_{0}} \left(1 - e\right)\).

Step by step solution

01

Express Volume Charge Density as Integral

Express the volume charge density, \(\rho\), as an integral over the volume, \(\frac{Q}{V}\), where \(Q\) is the total charge and \(V\) is the volume of the sphere. To find \(Q\), integrate the volume charge density over the volume of the sphere. Note that \(\rho_{0}e^{r / R}\) is the charge density and can be written as \(\frac{dQ}{dV}\). Therefore, it stands to \( dQ = \rho_{0}e^{r / R} dV \).
02

Calculate Total Charge of the Sphere

Next, calculate the total charge of the sphere by integrating both sides, resulting in \(Q = \rho_{0} \int e^{r / R} dV\). Use spherical coordinates to express \(dV\) as \(r^2 sin\theta dr d\phi d\theta\) and \(V\) as \(\int_0^R r^2 dr \int_0^{\pi}sin(\theta)d\theta \int_0^{2\pi}d\phi\). So \(Q = \rho_{0} \int_0^R e^{r / R} 4 \pi r^2 dr\).
03

Use Gauss’s Law to Find the Electric Field Strength

Finally, use Gauss’s law to find the electric field strength on the edge of the sphere, \(E\). Gauss’s law states that the electric field across a closed surface is given by the total charge enclosed divided by \(4 \pi \epsilon_0 r^2\), where \(\epsilon_0\) is the electric constant. So, \(E = \frac{Q}{4 \pi \epsilon_0 R^2}\). Substitute \(Q\) from Step 2 into the Gauss equation. After simplifying, we get the final equation.

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Most popular questions from this chapter

In Gauss's law, \(\oint \vec{E} \cdot d \vec{A}=q / \epsilon_{0},\) does the field \(\vec{E}\) necessarily arise only from charges within the closed surface?

If a charged particle were released from rest on a curved field line, would its subsequent motion follow the field line? Explain.

Why must the electric field be zero inside a conductor in electrostatic equilibrium?

If the flux of the gravitational field through a closed surface is zero, what can you conclude about the region interior to the surface?

A point charge of \(-2 Q\) is at the center of a spherical shell of radius \(R\) carrying charge \(Q\) spread uniformly over its surface. Find the electric field at (a) \(r=\frac{1}{2} R\) and (b) \(r=2 R\). (c) How would your answers change if the charge on the shell were doubled?
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