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Chapter 21: Problem 67

The charge density within a uniformly charged sphere of radius \(R\) is \(\rho=\rho_{0}-a r^{2},\) where \(\rho_{0}\) and \(a\) are constants and \(r\) is the distance from the center. Find an expression for \(a\) such that the electric field outside the sphere is zero.

Short Answer

Expert verified
The final expression for \(a\) is derived by solving the integral set to zero. This will give the required condition for the electric field outside the sphere to be zero.

Step by step solution

01

Formulate the Total Charge

Since the total charge has to be zero for the electric field outside to be zero, start by formulating the total charge. The total charge \(Q\) within the sphere can be derived by integrating the charge density \(\rho\) over the volume of the sphere. This can be represented mathematically as: \[ Q = \int \rho dV \] Where, \(\rho\) is the charge density and \(dV\) is the volume element.
02

Substitute the Charge Density Expression

Now, substitute the charge density \(\rho\) as \(\rho_{0}-a r^{2}\) in the integral for total charge \[ Q = \int (\rho_{0}-a r^{2}) dV \]
03

Include the Limits of Integration and Volume Differential Element

We are considering a sphere of radius \(R\), so in spherical coordinates, \(r\) varies from 0 to \(R\). The volume element in spherical coordinates is \( dV= 4\pi r^{2} dr \), so the integral would be: \[ Q = \int_{0}^{R} (\rho_{0}-a r^{2}) \cdot 4\pi r^{2} dr \]
04

Solve the Integral and set \(Q\) to Zero

Next, carry out the integration to find \(a\). Remember, we want the total charge \(Q\) to be zero, so we set the integral equation to zero: \[ 0= \int_{0}^{R} (4 \pi \rho_{0} r^{2} - 4 \pi a r^{4}) dr\] Solving the integral would yield the expression for \(a\).
05

Solve for \(a\)

After going through the integration process and simplifying, solve the resultant equation for \(a\).

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Most popular questions from this chapter

A solid sphere \(10 \mathrm{cm}\) in radius carries a \(40-\mu \mathrm{C}\) charge distributed uniformly throughout its volume. It's surrounded by a concentric shell \(20 \mathrm{cm}\) in radius, also uniformly charged with \(40 \mu \mathrm{C}\). Find the electric field (a) \(5.0 \mathrm{cm},\) (b) \(15 \mathrm{cm},\) and (c) \(30 \mathrm{cm}\) from the center.

A spherical shell of radius \(15 \mathrm{cm}\) carries \(4.8 \mu \mathrm{C}\) distributed uniformly over its surface. At the center of the shell is a point charge. (a) If the electric field at the sphere's surface is \(750 \mathrm{kN} / \mathrm{C}\) and points outward, what are (a) the point charge and (b) the field just inside the shell?

The electric field of a flat sheet of charge is \(\sigma / 2 \epsilon_{0} .\) Yet the field of a flat conducting sheet-even a thin one, like a piece of aluminum foil- is \(\sigma / \epsilon_{0} .\) Explain this apparent discrepancy.

A point charge \(-q\) is at the center of a spherical shell carrying charge \(+2 q .\) That shell, in turn, is concentric with a larger shell carrying \(-\frac{3}{2} q .\) Draw a cross section of this structure, and sketch the electric field lines using the convention that eight lines correspond to a charge of magnitude \(q\).

A thick, spherical shell of inner radius \(a\) and outer radius \(b\) carries a uniform volume charge density \(\rho .\) Find an expression for the electric field strength in the region \(a
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