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Chapter 22: Problem 59

A solid sphere of radius \(R\) has a nonuniform charge distribution \(\rho=A r^{2},\) where \(A\) is a constant. Determine the total charge, \(Q\), within the volume of the sphere.

Short Answer

Expert verified
Answer: The total charge (Q) within the sphere with the given nonuniform charge distribution is Q = (4πAR⁵)/5.

Step by step solution

01

Setup the integral

To find the total charge within the volume of the sphere, we need to integrate the charge distribution function \(\rho = Ar^2\) over the sphere of radius \(R\). We'll use the spherical coordinate system for the integral. We express the charge element \(dq\) as the product of the charge distribution function and the volume element \(dV\). In spherical coordinates, \(dV = r^2\sin\theta dr\, d\theta\, d\phi\). Therefore, the charge element can be expressed as \(dq = Ar^2(r^2\sin\theta dr\, d\theta\, d\phi)\).
02

Integrate over the volume

To find the total charge within the sphere, integrate the charge element \(dq\) over the entire volume. We have to integrate over \(r\), \(\theta\), and \(\phi\) separately: $$ Q = \int_{V}dq = \int_{r=0}^{R} \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} Ar^2(r^2\sin\theta dr\, d\theta\, d\phi). $$
03

Solve the integral

We'll solve the integral step by step, starting with \(\phi\): $$ Q = A \int_{r=0}^{R} r^4 dr \int_{\theta=0}^{\pi} \sin\theta\, d\theta \int_{\phi=0}^{2\pi} d\phi. $$ Now, we solve each integral one by one: $$ \int_{\phi=0}^{2\pi} d\phi = 2\pi, $$ $$ \int_{\theta=0}^{\pi} \sin\theta\, d\theta = -\cos\theta \Big|_0^\pi= 2, $$ and $$ \int_{r=0}^{R} r^4 dr= \frac{1}{5}r^5 \Big|_0^R = \frac{1}{5}R^5. $$ Plugging the results back into the expression for \(Q\), we get: $$ Q = A \cdot \frac{1}{5}R^5 \cdot 2 \cdot 2\pi. $$
04

Simplify the expression for the total charge

The final step is to simplify the expression for the total charge within the sphere. We have: $$ Q = \frac{4\pi AR^5}{5}. $$ Here, \(Q\) represents the total charge within the volume of the sphere with the nonuniform charge distribution \(\rho = Ar^2\).

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