### Electric Field Distribution of a Spherical Charge DistributionA uniformly charged spherical distribution is illustrated below. A spherical shell (blue) of radius \( b \) has volume charge density \( \rho_1 \) and the interior sphere (yellow) of radius \( a \) has volume charge density \( \rho_2 \). What is the electric field strength \( E \) outside the distribution for \( r > b \)? (Hint: use Gauss’s Law.)#### Diagram Explanation:The diagram presents a cross-section of two spherical regions:1. **Yellow Sphere (inner sphere)**: This sphere has a radius denoted as \( a \) and a volume charge density of \( \rho_2 \).2. **Blue Shell (outer shell)**: This shell extends from radius \( a \) to radius \( b \), with a volume charge density \( \rho_1 \).The black line passing through the spheres represents a radial distance \( r \) which is greater than the outer radius \( b \).In order to solve for the electric field \( E \) outside the spherical distribution (i.e., for \( r > b \)), we can employ Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by that surface.#### Gauss's Law Formula:\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]Based on this principle, considering a spherical Gaussian surface of radius \( r \) (where \( r > b \)):1. Calculate the total charge enclosed by the spherical distributions.2. Use the symmetry of the problem to ascertain that the electric field \( \vec{E} \) will be radially symmetric and constant over the Gaussian surface.To find the total charge \( Q_{\text{enc}} \):\[ Q_{\text{enc}} = \rho_2 \cdot \frac{4}{3} \pi a^3 + \rho_1 \cdot \left( \frac{4}{3} \pi b^3 - \frac{4}{3} \pi a^3 \right) \]Thereafter, using Gauss's Law:\[ E \cdot 4 \pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} \]By solving this equation, we Below is the transcribed text from the provided image, formatted as it would appear on an educational website:---### Multiple Choice Question**Instructions:** Choose the correct option from the following choices.#### Options:A. \(0\)B. \(\frac{d^2(p_1 - p_2) + b^2p_1}{3e_0b^2}\)C. \(\frac{d^2p_2}{3e_0b^2}\)D. \(\frac{d^2(p_1 + p_2) + b^2p_1}{3e_0b^2}\)E. \(\frac{d^2(p_1 - p_2) + b^2p_1}{3e_0b}\)---There are no graphs or diagrams included in this question. The question is written in a mathematical format and contains algebraic expressions. Make sure to carefully evaluate each option to find the correct solution.

Electric field strength:- Electric field strength is a vector quantity. It can be denoted by E. Unit…

A uniformly charged spherical distribution is illustrated below. A spherical shell (blue) of radius b has volume charge density and the interior sphere (yellow) of radius a has volume charge density p, What is the electric field strength E outside the distribution for r>b? (hint: use Gauss's _aw.) P₁ b a P2

A uniformly charged spherical distribution is illustrated below. A spherical shell (blue) of radius b has volume charge density and the interior sphere (yellow) of radius a has volume charge density p, What is the electric field strength E outside the distribution for r>b? (hint: use Gauss's _aw.) P₁ b a P2

Introductory Circuit Analysis (13th Edition)

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Publisher:Robert L. Boylestad

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Concept explainers

Spherical Coordinate System

The spherical coordinate system is a three-dimensional coordinate system for space where the position of a point is specified by three numbers- the radial distance of that point from a fixed source, its polar angle measured from a certain zenith, and the azimuth angle. Its orthogonal projection on the reference plane passes through the source and is measured from the direction of constant reference on that plane which is orthogonal to the highest level. It can be viewed as a three-dimensional version of the polar coordinate system.

Question

### Electric Field Distribution of a Spherical Charge Distribution

A uniformly charged spherical distribution is illustrated below. A spherical shell (blue) of radius \( b \) has volume charge density \( \rho_1 \) and the interior sphere (yellow) of radius \( a \) has volume charge density \( \rho_2 \). What is the electric field strength \( E \) outside the distribution for \( r > b \)? (Hint: use Gauss’s Law.)

#### Diagram Explanation:

The diagram presents a cross-section of two spherical regions:
1. **Yellow Sphere (inner sphere)**: This sphere has a radius denoted as \( a \) and a volume charge density of \( \rho_2 \).
2. **Blue Shell (outer shell)**: This shell extends from radius \( a \) to radius \( b \), with a volume charge density \( \rho_1 \).

The black line passing through the spheres represents a radial distance \( r \) which is greater than the outer radius \( b \).

In order to solve for the electric field \( E \) outside the spherical distribution (i.e., for \( r > b \)), we can employ Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by that surface.

#### Gauss's Law Formula:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]

Based on this principle, considering a spherical Gaussian surface of radius \( r \) (where \( r > b \)):
1. Calculate the total charge enclosed by the spherical distributions.
2. Use the symmetry of the problem to ascertain that the electric field \( \vec{E} \) will be radially symmetric and constant over the Gaussian surface.

To find the total charge \( Q_{\text{enc}} \):
\[ Q_{\text{enc}} = \rho_2 \cdot \frac{4}{3} \pi a^3 + \rho_1 \cdot \left( \frac{4}{3} \pi b^3 - \frac{4}{3} \pi a^3 \right) \]

Thereafter, using Gauss's Law:
\[ E \cdot 4 \pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} \]

By solving this equation, we

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