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<h2 class="hd hd-2 unit-title">W2PS1: Charge Density for Uniformly Charged Cylinder.</h2>
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Charge Density for Uniformly Charged Cylinder
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<p>
A solid cylinder, of length [mathjaxinline]L[/mathjaxinline] and radius [mathjaxinline]R[/mathjaxinline], is uniformly charged with total charge [mathjaxinline]Q[/mathjaxinline]. If needed, use pi for [mathjaxinline]\pi[/mathjaxinline] in your answers. </p>
<p><b class="bfseries">(Part a)</b> What is the <i class="itshape">volume charge density</i> [mathjaxinline]\rho[/mathjaxinline]? </p>
<p>
<p style="display:inline">[mathjaxinline]\rho =[/mathjaxinline]</p>
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</div><b class="bfseries">(Part b)</b> What is the <i class="itshape">linear charge density</i> [mathjaxinline]\lambda[/mathjaxinline]? </p>
<p>
<p style="display:inline">[mathjaxinline]\lambda =[/mathjaxinline]</p>
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<p><b class="bfseries">(Part c)</b> What is the relationship between [mathjaxinline]\rho[/mathjaxinline] and [mathjaxinline]\lambda[/mathjaxinline] (use lambda for [mathjaxinline]\lambda[/mathjaxinline])? </p>
<p>
<p style="display:inline">[mathjaxinline]\rho =[/mathjaxinline]</p>
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Continuous Charge Densities
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A cylindrical shell of length [mathjaxinline]L[/mathjaxinline] and radius [mathjaxinline]R[/mathjaxinline], with [mathjaxinline]L \gg R[/mathjaxinline], is uniformly charged on its surface with total charge [mathjaxinline]Q[/mathjaxinline]. We only place charge on the curved side surface of the cylinder. The end caps of the cylinder have no charge. </p>
<p><b class="bfseries">(Part a)</b> What is the <b class="bfseries"><i class="itshape">surface charge density</i></b> [mathjaxinline]\sigma[/mathjaxinline] on the cylinder in terms of the variables given? Check units! </p>
<p>
<p style="display:inline">[mathjaxinline]\sigma =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Suppose you go very far away from the cylinder to a distance much greater than [mathjaxinline]R[/mathjaxinline]. The cylinder now looks like a line of charge. What is the <b class="bfseries"><i class="itshape">linear charge density</i></b> [mathjaxinline]\lambda[/mathjaxinline] of that apparent line of charge? Check units! </p>
<p>
<p style="display:inline">[mathjaxinline]\lambda =[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">W2PS2: Electric Field of a Dipole Everywhere in Space.</h2>
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Electric field of a Dipole
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<h3>
<b class="bfseries">Introduction to the Electric Dipole</b>
</h3>
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<p>
An electric dipole consists of two equal but oppositely charged point-like objects, [mathjaxinline]+q[/mathjaxinline] and [mathjaxinline]-q[/mathjaxinline], separated by a distance [mathjaxinline]2a[/mathjaxinline], as shown in the figure below. Note that [mathjaxinline]\hat{i}[/mathjaxinline] and [mathjaxinline]\hat{j}[/mathjaxinline] are in the [mathjaxinline]+x[/mathjaxinline] and [mathjaxinline]+y[/mathjaxinline] directions, respectively, and [mathjaxinline]\hat{k}[/mathjaxinline] points out of the page. </p>
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<img src="/assets/courseware/v1/bbbf28ffc057da5952fc408229693c2f/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_FPS_01_Electric_Dipole-fig001.jpg" width="440"/>
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<p>
The dipole moment vector [mathjaxinline]\vec{\mathbf{p}}[/mathjaxinline] points from [mathjaxinline]-q[/mathjaxinline] to [mathjaxinline]+q[/mathjaxinline] and is given by (with [mathjaxinline]\hat{j}[/mathjaxinline] in the [mathjaxinline]+y[/mathjaxinline] direction): </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \vec{\mathbf{p}} =2qa\, \hat{\mathbf{j}}[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
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<p>
The magnitude of the electric dipole is [mathjaxinline]p=2qa[/mathjaxinline], where [mathjaxinline]q&gt;0[/mathjaxinline]. For an overall charge-neutral system having <i class="itshape">N</i> charged objects, the electric dipole vector [mathjaxinline]\vec{\mathbf{p}}[/mathjaxinline] is defined as </p>
<table id="a0000000004" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \vec{\mathbf{p}} \equiv \sum \limits _{i=1}^{i=N}q_{i} \vec{\mathbf{r}} _{\mathbf{i}}[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
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<p>
where [mathjaxinline]\vec{\mathbf{r}} _{\mathbf{i}}[/mathjaxinline] is the position vector of the charged object with charge [mathjaxinline]q_{i}[/mathjaxinline]. </p>
<p>
<b class="bfseries">Problem 1 Electric field of a Dipole</b>
</p>
<p>
Consider the electric dipole shown in the figure above. The electric fields of the two charges at the point [mathjaxinline]P[/mathjaxinline] located at [mathjaxinline](x,y,0)[/mathjaxinline], where [mathjaxinline]r=(x^{2} +y^{2} )^{1/2}[/mathjaxinline] and [mathjaxinline]\theta =\tan ^{-1} (x/y)[/mathjaxinline], are given by the expressions: </p>
<table id="a0000000006" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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[mathjaxinline]\displaystyle \vec{\mathbf{E}} _{+}[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle k_{e} \frac{q_{+} }{r_{+}^{3} } \vec{\mathbf{r}} _{+}[/mathjaxinline]
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<td style="width:20%; border:none" class="eqnnum">&#160;</td>
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<tr id="a0000000008">
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[mathjaxinline]\displaystyle \vec{\mathbf{E}} _{-}[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle k_{e} \frac{q_{-} }{r_{-}^{3} } \vec{\mathbf{r}} _{-}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
</tr>
</table>
<p><b class="bfseries">Question 1:</b> Find an expression for the vectors [mathjaxinline]\vec{\mathbf{r}} _{+}[/mathjaxinline] and [mathjaxinline]\vec{\mathbf{r}} _{-}[/mathjaxinline] in terms of [mathjaxinline]x[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], [mathjaxinline]a[/mathjaxinline], [mathjaxinline]\hat{\mathbf{i}}[/mathjaxinline] and [mathjaxinline]\hat{\mathbf{j}}[/mathjaxinline]. Write your answer using some or all of the following: x, y, a, 'hati' for [mathjaxinline]\hat{\mathbf{i}}[/mathjaxinline] and 'hatj' for [mathjaxinline]\hat{\mathbf{j}}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\vec{\mathbf{r}}_{+} =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\vec{\mathbf{r}}_{-} =[/mathjaxinline] </p>
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<p><b class="bfseries">Question 2:</b> Use the superposition principle to determine an expression for the [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]y[/mathjaxinline] components of the electric field at the point [mathjaxinline](x,y,0)[/mathjaxinline]. </p>
<p>
Note that we have added an explicit position of 0 in the [mathjaxinline]z[/mathjaxinline] direction. If the two charges and the point [mathjaxinline]P[/mathjaxinline] are all at [mathjaxinline]z=0[/mathjaxinline], the vectors [mathjaxinline]\vec{\mathbf{r}}_{+} =[/mathjaxinline] and [mathjaxinline]\vec{\mathbf{r}}_{-} =[/mathjaxinline] have no [mathjaxinline]\hat{k}[/mathjaxinline] component, and therefore the component of the electric fields in the [mathjaxinline]\hat{k}[/mathjaxinline] direction must also be zero. </p>
<table id="a0000000009" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000010">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \vec{\mathbf{E}} (x,y,0)=E_{x} (x,y,0)\hat{\mathbf{i}} +E_{y} (x,y,0)\hat{\mathbf{j}}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
</tr>
</table>
<p>
Express your answers as functions of the field point coordinates [mathjaxinline](x,y,0)[/mathjaxinline] using some or all of the following: k_e for [mathjaxinline]k_ e[/mathjaxinline], q, x, y, and a, as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]E_ x =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]E_ y =[/mathjaxinline] </p>
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We can show that the electric field of the dipole in the limit where [mathjaxinline]r&gt;&gt;a[/mathjaxinline] is </p>
<table id="a0000000011" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000012">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle E_{x} =\frac{k_{e} 3p}{r^{3} } \sin( \theta ) \cos( \theta ),\text { } E_{y} =\frac{k_{e} p}{r^{3} } (3\cos ^{2}( \theta )-1)[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
</tr>
</table>
<p>
where [mathjaxinline]\theta[/mathjaxinline] is the angle shown on the diagram so that [mathjaxinline]\sin (\theta ) =x/r[/mathjaxinline] and [mathjaxinline]\cos (\theta ) =y/r[/mathjaxinline]. <b class="bfseries">See Chapter 2.14.4 Electric Field of a Dipole.</b> </p>
<p><b class="bfseries">Question 3: </b>By what power of distance [mathjaxinline]r[/mathjaxinline] from the origin does the strength of the electric field fall off? How does this compare to a single point charge? Briefly explain a reason for the difference between these two cases. </p>
<p>
(There is no answer check for this question.) </p>
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<p><b class="bfseries">Note on coordinate system:</b> Since the dipole is cylindrically symmetric, if the point [mathjaxinline]P[/mathjaxinline] was not at [mathjaxinline]z=0[/mathjaxinline], we could always rotate our coordinate system so that the two charges and [mathjaxinline]P[/mathjaxinline] were all in the [mathjaxinline]x-y[/mathjaxinline] plane and get back to the situation shown above. </p>
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<h2 class="hd hd-2 unit-title">W2PS3: Electric Field of a charged Rod</h2>
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Continuous Charge Density: Set up and Integral - Charged rod - First Approach.
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<p>
A rod of length [mathjaxinline]L[/mathjaxinline] is uniformly charged with a total charge [mathjaxinline]Q[/mathjaxinline]. The electric field at point [mathjaxinline]P[/mathjaxinline] is expressed as: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \vec{E} = \int _{rod}\frac{kdq}{r_{dq,P}^2}\hat{r}_{dq,P}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
</td>
</tr>
</table>
<p>
where [mathjaxinline]\hat{r}_{dq,P}[/mathjaxinline] is the unit vector directed from the element of charge [mathjaxinline]dq[/mathjaxinline] in the rod to point [mathjaxinline]P[/mathjaxinline], and [mathjaxinline]r_{dq,P}[/mathjaxinline] is the distance from [mathjaxinline]dq[/mathjaxinline] to point [mathjaxinline]P[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> Consider the coordinate system shown in the figure with [mathjaxinline]x=0[/mathjaxinline] coinciding with the left end of the rod, [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]y[/mathjaxinline] are the components of point [mathjaxinline]P[/mathjaxinline], and [mathjaxinline]x_ s[/mathjaxinline] is the position of the element of charge [mathjaxinline]dq[/mathjaxinline]. Write an expression for [mathjaxinline]r_{dq,P}[/mathjaxinline], the distance between [mathjaxinline]dq[/mathjaxinline] and point [mathjaxinline]P[/mathjaxinline] in terms of [mathjaxinline]x[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], and x_s for [mathjaxinline]x_ s[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]r_{dq,P}=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part b)</b> The angle [mathjaxinline]\theta[/mathjaxinline] in the figure is the angle between [mathjaxinline]\hat{r}_{dq,P}[/mathjaxinline] and the horizontal. Write an expression for [mathjaxinline]\hat{r}_{dq,P}[/mathjaxinline], in terms of theta for [mathjaxinline]\theta[/mathjaxinline], hati for [mathjaxinline]\hat{i}[/mathjaxinline], and hatj for [mathjaxinline]\hat{j}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\hat{r}_{dq,P}=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part c)</b> Now, use the geometry shown to express the unit vector [mathjaxinline]\hat{r}_{dq,P}[/mathjaxinline] in terms of [mathjaxinline]x[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], x_s for [mathjaxinline]x_ s[/mathjaxinline], hati for [mathjaxinline]\hat{i}[/mathjaxinline], and hatj for [mathjaxinline]\hat{j}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\hat{r}_{dq,P}=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part d)</b> The electric field produced by the rod at point [mathjaxinline]P[/mathjaxinline] has two components and can be expressed as: </p>
<table id="a0000000008" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000009">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \vec{E} = \int _0^ Lf(x_ s) \; dx_ s\; \hat{i} + \int _0^ Lg(x_ s)\; dx_ s\; \hat{j}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
</tr>
</table>
<p>
Use the results obtained in parts (a) and (c) in equation (1) to find the functions [mathjaxinline]f(x_ s)[/mathjaxinline] and [mathjaxinline]g(x_ s)[/mathjaxinline] needed to calculate [mathjaxinline]\vec{E}[/mathjaxinline] at point [mathjaxinline]P[/mathjaxinline]. Express the answer in terms of [mathjaxinline]Q[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], [mathjaxinline]x[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], x_s for [mathjaxinline]x_ s[/mathjaxinline], and the Coulomb constant [mathjaxinline]k[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]f(x_ s) =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]g(x_ s)=[/mathjaxinline]</p>
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<h3 class="hd hd-2">W02PS01: Electric Field of a charged Rod</h3>
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<h2 class="hd hd-2 unit-title">W2PS4: Two charged circular arcs.</h2>
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Two Charged Circular Arcs
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<p>
Two circular arcs of radius [mathjaxinline]R[/mathjaxinline] are uniformly charged with a positvie charge per unit length [mathjaxinline]\lambda[/mathjaxinline]. The arcs lie on a plane as shown in the figure. Each arc subtends an angle [mathjaxinline]\theta =\pi /3=60^{\circ }[/mathjaxinline] . </p>
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<img src="/assets/courseware/v1/8b1ce5001baf9d26ed53ca128b3e2717/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_W2D3_arcs_01.svg" width="330"/>
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<p>
What is the direction and magnitude of the electric field anywhere along the [mathjaxinline]z[/mathjaxinline] axis that passes through the center of the circular arcs, perpendicular to the plane of the figure? </p>
<p>
Write the magnitude of the electric field using some or all of the following: [mathjaxinline]z[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], 'lambda' for [mathjaxinline]\lambda[/mathjaxinline] , 'pi' for [mathjaxinline]\pi[/mathjaxinline], 'epsilon_0' for [mathjaxinline]\varepsilon _0[/mathjaxinline] (where [mathjaxinline]k_ e=1/(4\pi \epsilon _0[/mathjaxinline]) as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]E(z)=[/mathjaxinline] </p>
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<div id="display_fps_fridayw2_3_sol_2_1" class="equation">`{::}`</div>
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<p>
For the direction of the electric field, specify which of the components ([mathjaxinline]\hat{x}[/mathjaxinline], [mathjaxinline]\hat{y}[/mathjaxinline], and/or [mathjaxinline]\hat{z}[/mathjaxinline]) are non-zero and for the non-zero cases, whether the field points in the plus or minus direction. Assume that the [mathjaxinline]y[/mathjaxinline] axis points in/out in the figure (i.e. it bisects the gaps between arcs) with [mathjaxinline]+y[/mathjaxinline] pointing away from you and the [mathjaxinline]x[/mathjaxinline] axis points left/right and bisects the two arcs (with [mathjaxinline]+x[/mathjaxinline] pointing to the right). </p>
<p>
Which components are non-zero? (check all that apply) <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_fps_fridayw2_3_sol_3_1">
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<input type="checkbox" name="input_fps_fridayw2_3_sol_3_1[]" id="input_fps_fridayw2_3_sol_3_1_choice_0" class="field-input input-checkbox" value="choice_0"/><label id="fps_fridayw2_3_sol_3_1-choice_0-label" for="input_fps_fridayw2_3_sol_3_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_fps_fridayw2_3_sol_3_1"> <text>[mathjaxinline]+\hat{x}[/mathjaxinline]</text>
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<input type="checkbox" name="input_fps_fridayw2_3_sol_3_1[]" id="input_fps_fridayw2_3_sol_3_1_choice_1" class="field-input input-checkbox" value="choice_1"/><label id="fps_fridayw2_3_sol_3_1-choice_1-label" for="input_fps_fridayw2_3_sol_3_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_fps_fridayw2_3_sol_3_1"> <text>[mathjaxinline]-\hat{x}[/mathjaxinline]</text>
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<input type="checkbox" name="input_fps_fridayw2_3_sol_3_1[]" id="input_fps_fridayw2_3_sol_3_1_choice_2" class="field-input input-checkbox" value="choice_2"/><label id="fps_fridayw2_3_sol_3_1-choice_2-label" for="input_fps_fridayw2_3_sol_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_fps_fridayw2_3_sol_3_1"> <text>[mathjaxinline]+\hat{y}[/mathjaxinline]</text>
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<input type="checkbox" name="input_fps_fridayw2_3_sol_3_1[]" id="input_fps_fridayw2_3_sol_3_1_choice_3" class="field-input input-checkbox" value="choice_3"/><label id="fps_fridayw2_3_sol_3_1-choice_3-label" for="input_fps_fridayw2_3_sol_3_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_fps_fridayw2_3_sol_3_1"> <text>[mathjaxinline]-\hat{y}[/mathjaxinline]</text>
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<input type="checkbox" name="input_fps_fridayw2_3_sol_3_1[]" id="input_fps_fridayw2_3_sol_3_1_choice_4" class="field-input input-checkbox" value="choice_4"/><label id="fps_fridayw2_3_sol_3_1-choice_4-label" for="input_fps_fridayw2_3_sol_3_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_fps_fridayw2_3_sol_3_1"> <text>[mathjaxinline]+\hat{z}[/mathjaxinline]</text>
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<input type="checkbox" name="input_fps_fridayw2_3_sol_3_1[]" id="input_fps_fridayw2_3_sol_3_1_choice_5" class="field-input input-checkbox" value="choice_5"/><label id="fps_fridayw2_3_sol_3_1-choice_5-label" for="input_fps_fridayw2_3_sol_3_1_choice_5" class="response-label field-label label-inline" aria-describedby="status_fps_fridayw2_3_sol_3_1"> <text>[mathjaxinline]-\hat{z}[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">W2PS5: Electric Field of a uniform charged disk.</h2>
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<h3 class="hd hd-3 problem-header" id="fps_fridayw2_5-problem-title" aria-describedby="block-v1:MITx+8.02.1x+1T2019+type@problem+block@fps_fridayw2_5-problem-progress" tabindex="-1">
Electric Field Due to Uniformly Charged Disk
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<img src="/assets/courseware/v1/31c94b0522d246245b0db0b29f4f426d/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_friday2_fig051.svg" width="330"/>
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<p>
A disc of radius [mathjaxinline]R[/mathjaxinline] is uniformly charged with total charge [mathjaxinline]Q &gt; 0[/mathjaxinline]. Determine the direction and magnitude of the electric field at the point [mathjaxinline]P[/mathjaxinline] lying a distance [mathjaxinline]x &gt; 0[/mathjaxinline] from the center of the disc along the axis of symmetry of the disc. </p>
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We will build up the electric field of the disc by integrating the contributions from the electric field due to a series of rings of radius [mathjaxinline]r[/mathjaxinline], thickness [mathjaxinline]dr[/mathjaxinline], and charge [mathjaxinline]dq[/mathjaxinline]. Thus the radius of the ring [mathjaxinline]r[/mathjaxinline] becomes our integration variable. </p>
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<img src="/assets/courseware/v1/4e40c294bd90b0751cdc3a26be66bc64/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_friday2_fig052New.svg" width="330"/>
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<p>
In worked example L04v05, it was shown that the electric field of a ring of radius [mathjaxinline]r[/mathjaxinline] points along the [mathjaxinline]x[/mathjaxinline] axis and has the following magnitude (note that in the lesson, the charge on the ring was calculated from the linear charge density [mathjaxinline]q_{ring}=\lambda (2\pi r)[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] was used instead of [mathjaxinline]1/4\pi \epsilon _0[/mathjaxinline]): </p>
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[mathjaxinline]\displaystyle dE_ x = {1 \over {4 \pi \varepsilon _0}} {{dq\ x} \over {\left( x^2 + r^2 \right) ^{3/2}}}[/mathjaxinline]
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<p><b class="bfseries">Question 1:</b> Determine an expression for [mathjaxinline]dq[/mathjaxinline] in terms of the integration variable [mathjaxinline]r[/mathjaxinline], the width of the ring [mathjaxinline]dr[/mathjaxinline], the total charge on the disk [mathjaxinline]Q[/mathjaxinline], and the disk radius [mathjaxinline]R[/mathjaxinline]. Write your answer using some or all of the following: Q, R, r, and dr. </p>
<p>
<p style="display:inline">[mathjaxinline]dq =[/mathjaxinline] </p>
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<p><b class="bfseries">Question 2:</b> Integrate your result to find an expression for the electric field of the disc. You may find the following integral formula useful: </p>
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[mathjaxinline]\displaystyle \int \limits _{r=0}^ R {rdr \over {\left( x^2 + r^2 \right) ^{3/2}}} = - \left. {1 \over {\left( x^2 + r^2 \right) ^{1/2}}}\right|_{r=0}^{r=R}=-\frac{1}{\left( x^2 + R^2 \right)^{1/2}}+\frac{1}{x}[/mathjaxinline]
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Write your answer using some or all of the following: Q, R, x, 'pi' for [mathjaxinline]\pi[/mathjaxinline], 'epsilon_0' for [mathjaxinline]\varepsilon _0[/mathjaxinline], 'hati' for [mathjaxinline]\hat{\mathbf{i}}[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\vec{\mathbf{E}}_{\text {disk}} =[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_fps_fridayw2_5_3_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">Question 3:</b> You should also check the limits. What is the limit of your expression for the electric field of the disc for the case that (i) [mathjaxinline]R \to 0[/mathjaxinline] (but keep [mathjaxinline]Q[/mathjaxinline] constant!), and for (ii) [mathjaxinline]R \to \infty[/mathjaxinline] such that [mathjaxinline]Q/(\pi R^2)[/mathjaxinline] remains constant? </p>
<p>
Hint: For [mathjaxinline]R \ll x[/mathjaxinline], use the power series expansion: </p>
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[mathjaxinline]\displaystyle {1 \over {\left( x^2 + R^2 \right)^{1/2}}} = {1 \over {x \left( 1 + \left( R\, / \, x \right)^2 \right)^{1/2}}} \simeq {1 \over x} \left( 1 - {1 \over 2} \left( R\, / \, x \right)^2 + \cdots \right)[/mathjaxinline]
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Write your answer using some or all of the following: Q, R, x, 'pi' for [mathjaxinline]\pi[/mathjaxinline], 'epsilon_0' for [mathjaxinline]\varepsilon _0[/mathjaxinline], 'hati' for [mathjaxinline]\hat{\mathbf{i}}[/mathjaxinline]. </p>
<p><b class="bfseries">(Part i)</b> When [mathjaxinline]R \to 0[/mathjaxinline], the electric field approaches <p style="display:inline">[mathjaxinline]\vec{\mathbf{E}}_{\text {disk}} =[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_fps_fridayw2_5_4_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part ii)</b> When [mathjaxinline]R \to \infty[/mathjaxinline], the electric field approaches <p style="display:inline">[mathjaxinline]\vec{\mathbf{E}}_{\text {disk}} =[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 4" role="group"><div id="inputtype_fps_fridayw2_5_5_1" class="text-input-dynamath capa_inputtype inline textline">
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