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Chapter 22: Problem 65

A sphere of radius \(R\) carries a nonuniform but spherically symmetric volume charge density that results in an electric field in the sphere given by \(\vec{E}=E_{0}(r / R)^{2} \hat{r},\) where \(E_{0}\) is a constant. Find the potential difference from the sphere's surface to its center.

Short Answer

Expert verified
The potential difference from the sphere's surface to its center is \(\frac{1}{3} E_{0} R\).

Step by step solution

01

Write down the formula for voltage

Potential difference, \(V\), is given by the negative integral of the electric field over a path: \(V = - \int_{a}^{b} \vec{E} \cdot d\vec{l}\), where \(a\) is the starting point, \(b\) is the ending point, and \(d\vec{l}\) is the infinitesimal path element. In this problem, we want to find the potential difference from the sphere's surface, \(R\), to its center, \(0\).
02

Set up the path integral

Since the electric field is directed radially (\(\hat{r}\)) and the path from the sphere's surface to the center is also radial, \(d\vec{l} = -dr\hat{r}\), gives a scalar equation: \(V = - \int_{R}^{0} E_{0}(r / R)^{2} (-dr)\). The extra negative sign is because the path is from the surface to the center. It comes from that the element \(dr\) goes in opposite direction to \(\hat{r}\).
03

Solve the integral

Executing the integral yields \(V = E_{0} \int_{R}^{0} (r / R)^{2} dr = [E_{0}(r^{3}/(3R^{2})]_{R}^{0} = \frac{1}{3} E_{0} R\).

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Most popular questions from this chapter

Show that the result of Example 22.8 approaches the field of a point charge for \(x \gg a\). (Hint: You'll need to apply the binomial approximation from Appendix A to the expression \(1 / \sqrt{x^{2}+a^{2}}\) )

Two points \(A\) and \(B\) lie \(15 \mathrm{cm}\) apart in a uniform electric field, with the path \(A B\) parallel to the field. If the potential difference \(\Delta V_{A B}\) is \(840 \mathrm{V},\) what's the field strength?

Find the potential as a function of position in the electric field \(\vec{E}=a x \hat{\imath},\) where \(a\) is a constant and \(V=0\) at \(x=0\)

Show that \(1 \mathrm{V} / \mathrm{m}\) is the same as \(1 \mathrm{N} / \mathrm{C}\)

The classical picture of the hydrogen atom has the electron orbiting 0.0529 nm from the proton. What's the electric potential associated with the proton's electric field at this distance?
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