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Q11P

Expert-verifiedFound in: Page 679

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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 ${{\mathit{x}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{5.00}}{\mathbf{}}{\mathbf{\text{m}}}$****, ${{\mathbf{y}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.00}}{\mathbf{\text{m}}}$****. ****The cube lies in a region where the electric field vector is given by $\overrightarrow{\mathbf{E}}{\mathbf{=}}\mathbf{[}\mathbf{-}\mathbf{3}\mathbf{.00}\mathbf{}\widehat{\mathbf{i}}\mathbf{-}\mathbf{4}\mathbf{.00}{\mathbf{y}}^{\mathbf{2}}\widehat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.00}\widehat{\mathbf{k}}\mathbf{]}{\mathbf{}}{\mathbf{\text{N/C}}}$****with y in meters. What is the net charge contained by the cube?**

** **

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

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

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

**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)

** **

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

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

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:

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

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:

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

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

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

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

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

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

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