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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 = (4蟺AR鈦�)/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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Most popular questions from this chapter

Two parallel, uniformly charged, infinitely long wires carry opposite charges with a linear charge density \(\lambda=1.00 \mu \mathrm{C} / \mathrm{m}\) and are \(6.00 \mathrm{~cm}\) apart. What is the magnitude and direction of the electric field at a point midway between them and \(40.0 \mathrm{~cm}\) above the plane containing the two wires?

A solid nonconducting sphere has a volume charge distribution given by \(\rho(r)=(\beta / r) \sin (\pi r / 2 R) .\) Find the total charge contained in the spherical volume and the electric field in the regions \(rR\). Show that the two expressions for the electric field equal each other at \(r=R\).

A spherical aluminized Mylar balloon carries a charge \(Q\) on its surface. You are measuring the electric field at a distance \(R\) from the balloon's center. The balloon is slowly inflated, and its radius approaches but never reaches R. What happens to the electric field you measure as the balloon increases in radius. Explain.

A uniformly charged rod of length \(L\) with total charge \(Q\) lies along the \(y\) -axis, from \(y=0\) to \(y=L\). Find an expression for the electric field at the point \((d, 0)\) (that is, the point at \(x=d\) on the \(x\) -axis).

A solid metal sphere of radius \(8.00 \mathrm{~cm},\) with a total charge of \(10.0 \mu C\), is surrounded by a metallic shell with a radius of \(15.0 \mathrm{~cm}\) carrying a \(-5.00 \mu \mathrm{C}\) charge. The sphere and the shell are both inside a larger metallic shell of inner radius \(20.0 \mathrm{~cm}\) and outer radius \(24.0 \mathrm{~cm} .\) The sphere and the two shells are concentric. a) What is the charge on the inner wall of the larger shell? b) If the electric field outside the larger shell is zero, what is the charge on the outer wall of the shell?
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