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    <name>
      <text>9.3.5.2 Computing the electric field</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p class="HELM_exercise">Exercise</p><p>Eight point charges of \({@charge_fact@}\ \mathrm{nC}\) each are located at the corners of a cube in free space which is \(1\ \mathrm{m}\) on each side (see the figure below). Calculate <span class="math inline">\(|{\underline E}|\)</span> at</p><p><br></p>
<ol class="HELM_parts">

<li><p>the centre of the cube: [[input:ans1]] \(\mathrm{V}\,\mathrm{m}^{-1}\) [[validation:ans1]] [[feedback:prt1]]</p></li>

<li><p>the centre of any face: [[input:ans2]] \(\mathrm{V}\,\mathrm{m}^{-1}\) [[validation:ans2]] [[feedback:prt2]]</p></li>

<li><p>the centre of any edge: [[input:ans3]] \(\mathrm{V}\,\mathrm{m}^{-1}\) [[validation:ans3]] [[feedback:prt3]]</p></li>
</ol>

<p><b>Note: Please write your answers with 2 decimal places</b></p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/Fig%280%292.svg" alt="sketch of cube" role="presentation" class="img-responsive atto_image_button_text-bottom" width="229" height="199">
</figure>]]></text>
<file name="Fig(0)2.svg" path="/" encoding="base64">[encoded data, 49920 characters]</file>
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      <text><![CDATA[<ol class="HELM_parts">

<li><p>You should know that the field at the centre of the cube is zero because of the symmetrical distribution of the charges.</p></li>

<li><p>Because of the symmetrical nature of the problem it does not matter which face is chosen in order to find the magnitude of the field at the centre of a face. Suppose the chosen face has corners located at \(P(1,1,1), T(1,1,0), R(0,1,0)\) and \(S(0,1,1)\) then the centre (\(C\)) of this face can be seen from the diagram to be located at \(C\left(\dfrac{1}{2}, 1,\dfrac{1}{2}\right)\).</p>
<p>The electric field at \(C\) due to the charges at the corners \(P,T,R\) and \(S\) will then be zero since the field vectors due to equal charges located at opposite corners of the square \(PTRS\) cancel one another out. The field at \(C\) is then due to the equal charges located at the remaining four corners \((OABD)\) of the cube, and we note from the symmetry of the cube, that the distance of each of these corners from \(C\) will be the same. In particular the distance </p>
\[OC=\sqrt{\left(\dfrac{1}{2}\right)^2+1^2+\left(\dfrac{1}{2}\right)^2}=\sqrt{1.5}\ \text{m}.\]
<p>The electric field \(\underline{E}\) at \(C\) due to the remaining charges can then be found using \(\underline{E}=\dfrac{1}{4\pi\varepsilon_0}\sum_1^4\dfrac{q_i\cdot \underline{r}_i}{|\underline{r}_i|^3}\) where \(q_1\) to \(q_4\) are the equal charges (\({@charge_fact@} \times 10^{-9}\) coulombs) and \(\underline{r}_1\) to \(\underline{r}_4\) are the vectors directed from the four corners, where the charges are located, towards \(C\). In this case since \(q_1={@charge_fact@} \times 10^{-9}\) coulombs and \(|\underline{r}_i|=\sqrt{1.5}\) for \(i=1\) to \(i=4\) we have</p><p><br></p><p>\(\underline{E}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{{@charge_fact@} \times 10^{-9}}{(1.5)^{3/2}}\left[\underline{r}_1+\underline{r}_2+\underline{r}_3+\underline{r}_4\right],\)</p><p><br></p><p>where \(\underline{r}_1=\underline{AC}=\left[\begin{array}{c}\dfrac{1}{2}\\ 1\\ \dfrac{1}{2}\end{array}\right]-\left[\begin{array}{c}0\\0\\1\end{array}\right],\ \underline{r}_2=\underline{BC} =\left[\begin{array}{c}\dfrac{1}{2}\\ 1\\ \dfrac{1}{2}\end{array}\right]-\left[\begin{array}{c}1\\0\\1\end{array}\right] \text{etc.}\)</p><p><br></p><p>Thus \(\underline{E}=\dfrac{1}{4\pi \varepsilon_0} \dfrac{{@charge_fact@} \times 10^{-9}}{(1.5)^{3/2}} \left[\begin{array}{c}0\\4\\0\end{array}\right]\)</p><p><br></p><p>and \(|\underline{E}|=\dfrac{1}{\pi\varepsilon_0}\ \dfrac{{@charge_fact@} \times 10^{-9}}{(1.5)^{3/2}}=\dfrac{{@charge_fact@} \times 10^{-9}}{\pi\times 8.854\times 10^{-12}(1.5)^{3/2}} ={@ta_2@}\ \mathrm{V}\,\mathrm{m}^{-1}\).</p></li>

<li><p>Suppose the chosen edge to be used connects \\(A(0,0,1)\\) to \\(B(1,0,1)\\) then the centre point \\((G)\\) will be located at \\(G\left(\dfrac{1}{2}, 0, 1\right)\\).</p><p><br></p><p>By symmetry, the field at \(G\) due to the charges at \(A\) and \(B\) will be zero.</p><p><br></p><p>We note that the distances \(DG, \ OG, \ PG\) and \(SG\) are all equal. In the case of \(OG\) we calculate by Pythagoras that this distance is \(\sqrt{\left(\dfrac{1}{2}\right)^2+0^2+1^2}=\sqrt{1.25}\).</p><p><br></p><p>Similarly the distances \(TG\) and \(RG\) are equal to \(\sqrt{2.25}\).</p><p><br></p><p>Using the result that \(\underline{E}=\dfrac{1}{4\pi \varepsilon_0} \sum \dfrac{q_i\underline{r}_i}{|\underline{r}_i|^3}\) gives</p><p><br></p><p>\(\qquad \underline{E}=\dfrac{{@charge_fact@} \times 10^{-9}}{4\pi\varepsilon_0}\left[\dfrac{1}{(1.25)^{3/2}}\left\{\left[\begin{array}{c}-\dfrac{1}{2}\\0\\1\end{array}\right]+\left[\begin{array}{c}\dfrac{1}{2}\\0\\1\end{array}\right]+\left[\begin{array}{c}-\dfrac{1}{2}\\-1\\0\end{array}\right]+\left[\begin{array}{c}\dfrac{1}{2}\\-1\\0\end{array}\right]\right\}\right.\)</p><p><br></p><p>\(\qquad\qquad\qquad\left.+\dfrac{1}{(2.25)^{3/2}}\left\{\left[\begin{array}{c}-\dfrac{1}{2}\\-1\\1\end{array}\right]+\left[\begin{array}{c}\dfrac{1}{2}\\-1\\1\end{array}\right]\right\}\right]\)</p><p><br></p><p>\(\qquad\quad =\dfrac{{@charge_fact@} \times 10^{-9}}{4\pi\varepsilon_0}\left[\dfrac{1}{(1.25)^{3/2}}\left[\begin{array}{c}0\\-2\\2\end{array}\right]+\dfrac{1}{(2.25)^{3/2}}\left[\begin{array}{c}0\\-2\\2\end{array}\right]\right]\)</p><p><br></p><p>\(\qquad\quad =\dfrac{{@charge_fact@} \times 10^{-9}}{4\pi \varepsilon_0}\left[\begin{array}{c}0\\-2.02367\\2.02367\end{array}\right]\)</p><p><br></p><p>Thus \(|\underline{E}|=\dfrac{{@charge_fact@} \times 10^{-9}}{4\times \pi\times 8.854\times 10^{-12}} \sqrt{0^2+(-2.02367)^2+(2.02367)^2}\)</p><p><br></p><p>\(\qquad\qquad ={@ta_3@}\ \mathrm{V}\,\mathrm{m}^{-1}\) (2 d.p.).</p></li>

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      <text>charge_fact: 1 + rand(8);
ta_2: decimalplaces(charge_fact*19.57,2);
ta_3: decimalplaces(charge_fact*25.72,2);</text>
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      <text>\(|{\underline E}|\) at the centre of any face: \( {@ta_2@} \)
\(|{\underline E}|\) at the centre of any edge: \( {@ta_3@} \)</text>
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