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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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Most popular questions from this chapter

Two infinite, uniformly charged, flat nonconducting surfaces are mutually perpendicular. One of the surfaces has a charge distribution of \(+30.0 \mathrm{pC} / \mathrm{m}^{2}\), and the other has a charge distribution of \(-40.0 \mathrm{pC} / \mathrm{m}^{2}\). What is the magnitude of the electric field at any point not on either surface?

Two uniformly charged insulating rods are bent in a semicircular shape with radius \(r=10.0 \mathrm{~cm} .\) If they are positioned so they form a circle but do not touch and have opposite charges of \(+1.00 \mu \mathrm{C}\) and \(-1.00 \mu \mathrm{C}\) find the magnitude and direction of the electric field at the center of the composite circular charge configuration.

A charge per unit length \(+\lambda\) is uniformly distributed along the positive \(y\) -axis from \(y=0\) to \(y=+a\). A charge per unit length \(-\lambda\) is uniformly distributed along the negative \(y\) axis from \(y=0\) to \(y=-a\). Write an expression for the electric field (magnitude and direction) at a point on the \(x\) -axis a distance \(x\) from the origin.

A point charge, \(+Q\), is located on the \(x\) -axis at \(x=a\), and a second point charge, \(-Q\), is located on the \(x\) -axis at \(x=-a\). A Gaussian surface with radius \(r=2 a\) is centered at the origin. The flux through this Gaussian surface is a) zero. b) greater than zero. c) less than zero. d) none of the above.

Two charges, \(+e\) and \(-e,\) are a distance of \(0.68 \mathrm{nm}\) apart in an electric field, \(E,\) that has a magnitude of \(4.4 \mathrm{kN} / \mathrm{C}\) and is directed at an angle of \(45^{\circ}\) with the dipole axis. Calculate the dipole moment and thus the torque on the dipole in the electric field.
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