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Q11P

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Found in: Page 679

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# Figure 23-35 shows a closed Gaussian surface in the shape of a cube of edge length ${\mathbf{2}}{\mathbf{.00}}{\mathbf{}}{\mathbf{\text{m}}}{\mathbf{,}}$ with ONE corner at ${{\mathbit{x}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{5.00}}{\mathbf{}}{\mathbf{\text{m}}}$, ${{\mathbf{y}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.00}}{\mathbf{\text{m}}}$. The cube lies in a region where the electric field vector is given by $\stackrel{\mathbf{\to }}{\mathbf{E}}{\mathbf{=}}\mathbf{\left[}\mathbf{-}\mathbf{3}\mathbf{.00}\mathbf{}\stackrel{\mathbf{^}}{\mathbf{i}}\mathbf{-}\mathbf{4}\mathbf{.00}{\mathbf{y}}^{\mathbf{2}}\stackrel{\mathbf{^}}{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.00}\stackrel{\mathbf{^}}{\mathbf{k}}\mathbf{\right]}{\mathbf{}}{\mathbf{\text{N/C}}}$with y in meters. What is the net charge contained by the cube?

The net charge contained by the cube is $-1.70×{10}^{-9}\text{C}$.

See the step by step solution

## Step 1: The given data

The electric field, $\stackrel{\to }{\mathrm{E}}=\left[-3.00\stackrel{^}{\mathrm{i}}-4.00{\mathrm{y}}^{2}\stackrel{^}{\mathrm{j}}+3.00\stackrel{^}{\mathrm{k}}\right]\text{\hspace{0.17em}N/C}$

The edge length of the cube is $\mathrm{a}=2.00\text{\hspace{0.17em}m}$ with one corner at ${\mathrm{x}}_{1}=5.00\text{\hspace{0.17em}m}$ , ${\mathrm{y}}_{1}=4.00\text{\hspace{0.17em}m}$

## Step 2: Understanding the concept of Gauss law-planar symmetry

Using the gauss flux theorem, we can get the net flux through the surfaces. Now, using the same concept, we can get the net charge contained in the cube.

Formula:

The electric flux passing through the surface,

$\mathrm{\varphi }=\mathrm{E}\cdot \mathrm{A}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{0}}$ (1)

## Step 3: Calculation of the net charge

None of the constant terms will result in a nonzero contribution to the flux, so we focus on the x-dependent term only:

${\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}=-4.00{\mathrm{y}}^{2}\stackrel{^}{\mathrm{i}}$

The face of the cube located at $\mathrm{y}=4.00\text{\hspace{0.17em}}\mathrm{m}$ has an area $\mathrm{A}=4.00\text{\hspace{0.17em}}{\mathrm{m}}^{2}$ (and it “faces” the +j direction) and has a “contribution” to the flux that is given using equation (1) as:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}\mathrm{A}=\left(3.00\text{\hspace{0.17em}}\mathrm{N}/\mathrm{C}\right)\left(-4.00×{\left(4.00\text{\hspace{0.17em}}\mathrm{m}\right)}^{2}\right)\\ =-256\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C}\end{array}$

The face of the cube located at $\mathrm{y}=2.00\text{\hspace{0.17em}}\mathrm{m}$ has the same area A (however, this one “faces” the –j direction) and a contribution to the flux that is given using equation (1) as:

$\begin{array}{c}-{\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}\mathrm{A}=\left(-4.00\text{\hspace{0.17em}}\mathrm{N}/\mathrm{C}\right)\left(-4.00×{\left(2.00\text{\hspace{0.17em}}\mathrm{m}\right)}^{2}\right)\\ =64\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C}\end{array}$

Thus, the net flux is given using equations (a) and (b) as given:

$\begin{array}{c}\mathrm{\varphi }=\left(-256+64\right)\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C}\\ =-192\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C}\end{array}$ .

According to Gauss’s law, the net charge contained by the face is given as:

$\begin{array}{c}{\mathrm{q}}_{\mathrm{enc}}=\left(8.85×{10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)\left(-192\text{\hspace{0.17em}}\text{N}\cdot {\text{m}}^{\text{2}}\text{/C}\right)\\ =-1.70×{10}^{-9}\text{C}\end{array}$

Hence, the value of the charge is $-1.70×{10}^{-9}\text{C}$.

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