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<h2 class="hd hd-2 unit-title">W4PS1: Potential for an Electric Dipole</h2>
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Electric Potential for Dipole
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<p>
Consider two oppositely charge objects located along the [mathjaxinline]y[/mathjaxinline]-axis. The positively charged object has charge [mathjaxinline]+q[/mathjaxinline] and is located at [mathjaxinline]y=+a[/mathjaxinline], and the negatively charged object has charge [mathjaxinline]-q[/mathjaxinline] and is located at [mathjaxinline]y=-a[/mathjaxinline], forming an electric dipole along the <i class="itshape">y</i>-axis, as shown in the figure below. </p>
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<img src="/assets/courseware/v1/7a28868c80c2cc1269b1791effbeb3d6/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_ps03_v02-fig042.png" width="400"/>
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<p><b class="bfseries">(Part a)</b> Use polar coordinates [mathjaxinline](r,\theta )[/mathjaxinline] for the point [mathjaxinline]P[/mathjaxinline] in the <i class="itshape">xy</i>-plane, find an expression for the electric potential [mathjaxinline]V(r,\theta )[/mathjaxinline] at the point [mathjaxinline]P[/mathjaxinline] assuming [mathjaxinline]V(\infty )=0[/mathjaxinline]. </p>
<p>
For the symbolic check, write your answer using some or all of the following: [mathjaxinline]q[/mathjaxinline], [mathjaxinline]a[/mathjaxinline], [mathjaxinline]r[/mathjaxinline], theta for [mathjaxinline]\theta[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] for [mathjaxinline]k \equiv 1/(4 \pi \varepsilon _0)[/mathjaxinline]. Use cos(), sin() and sqrt() as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]V(r, \theta ) =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> When [mathjaxinline]r \gg a[/mathjaxinline], use a Taylor series expansion to show that the electric potential can be approximated by the expression [mathjaxinline]V(r,\theta )= k \vec{\textbf{p}} \cdot \hat{\textbf{r}} / r^{2}[/mathjaxinline] where [mathjaxinline]\vec{\textbf{p}}[/mathjaxinline] is the electric dipole moment and [mathjaxinline]\hat{\textbf{r}}[/mathjaxinline] is the radial unit vector in polar coordinates. </p>
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<p><b class="bfseries">(Part c)</b> The component of the electric field in polar coordinates are given by the expressions </p>
<table id="a0000000006" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
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<td class="equation" style="width:80%; border:none">[mathjax]\begin{eqnarray*} E_{\theta } (r,\theta ) &amp; =&amp; -\frac{1}{r} \frac{\partial }{\partial \theta } V(r,\theta ) \\ E_{r} (r,\theta ) &amp; =&amp; -\frac{\partial }{\partial r} V(r,\theta ). \end{eqnarray*}[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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</table>
<p>
Calculate the components of the electric field in polar coordinates at the point [mathjaxinline]P[/mathjaxinline] associated with your &#8220;dipole approximation" in <b class="bfseries">(Part b)</b>. </p>
<p>
For the symbolic check, write your answer using some or all of the following: [mathjaxinline]p[/mathjaxinline], [mathjaxinline]r[/mathjaxinline], theta for [mathjaxinline]\theta[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] for [mathjaxinline]k \equiv 1/(4 \pi \varepsilon _0)[/mathjaxinline]. Use cos(), sin() and sqrt() as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]E_ r =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]E_\theta =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part d)</b> A positively charged dust particle with mass <i class="itshape">m</i> and charge [mathjaxinline]+q[/mathjaxinline] is released from rest at point B (not shown on the drawing) located on the [mathjaxinline]y[/mathjaxinline]-axis at the point given in polar coordinates by [mathjaxinline](d, 0)[/mathjaxinline], [mathjaxinline]d&gt;a[/mathjaxinline]. In what direction will it accelerate? What is the speed of the particle when it has traveled a distance [mathjaxinline]s[/mathjaxinline] from its original position at point B? Answer this question using the exact formulas, not the approximation given in Part b. </p>
<p>
For the symbolic check, write your answer using some or all of the following: [mathjaxinline]a[/mathjaxinline], [mathjaxinline]d[/mathjaxinline], [mathjaxinline]s[/mathjaxinline], [mathjaxinline]q[/mathjaxinline], [mathjaxinline]m[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] for [mathjaxinline]k \equiv 1/(4 \pi \varepsilon _0)[/mathjaxinline]. Use sqrt() as necessary. </p>
<p>
<p style="display:inline">[mathjaxinline]v =[/mathjaxinline] </p>
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<h3 class="hd hd-2">W04PS01a: Potential of a Dipole</h3>
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<h3 class="hd hd-2">W04PS01b: Potential of a Dipole - Full Expression</h3>
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<h2 class="hd hd-2 unit-title">W4PS2: Charged Particle Near a Uniform Ring</h2>
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Electric Potential for a Ring of Charge
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<p>
Consider a thin ring of charge with a total charge of [mathjaxinline]+Q[/mathjaxinline] and radius [mathjaxinline]R[/mathjaxinline] as shown in the figure below. </p>
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<p><b class="bfseries">(Part a)</b> Find an expression for the electric potential [mathjaxinline]V(z)[/mathjaxinline] at the point [mathjaxinline]P[/mathjaxinline] shown in the figure which is along the axis of the ring and a distance [mathjaxinline]z[/mathjaxinline] above its center. Use the convention where [mathjaxinline]V(\infty )=0[/mathjaxinline]. </p>
<p>
For the symbolic check, write your answer using some or all of the following: [mathjaxinline]Q[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]z[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline] for [mathjaxinline]k \equiv 1/(4 \pi \varepsilon _0)[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]V(z) =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> A small negatively charged particle with mass [mathjaxinline]m[/mathjaxinline] and charge [mathjaxinline]-q[/mathjaxinline] (with [mathjaxinline]q&gt;0)[/mathjaxinline] is released from rest at the point [mathjaxinline]P[/mathjaxinline]. In what direction will it accelerate? What is the speed of the particle when it reaches the center of the ring? </p>
<p>
For the symbolic check, write your answer using some or all of the following: [mathjaxinline]R[/mathjaxinline], [mathjaxinline]z[/mathjaxinline], [mathjaxinline]q[/mathjaxinline], [mathjaxinline]Q[/mathjaxinline], [mathjaxinline]m[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] for [mathjaxinline]k \equiv 1/(4 \pi \varepsilon _0)[/mathjaxinline]. Use sqrt() as necessary. </p>
<p>
<p style="display:inline">[mathjaxinline]v =[/mathjaxinline] </p>
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<p> Note that in the video below, the symbol "q" is used to denote a negative quantity, while in the problem above "q" is positive and the negative charge is denoted "-q".</p>
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<h3 class="hd hd-2">W04PS02: Speed of a Charged Particle Released Above a Ring</h3>
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<h2 class="hd hd-2 unit-title">W4PS3: Charged Disk</h2>
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Charged Disk
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<p>
In the presence of some distribution of charge, there are two distinct ways to find the electric potential difference between any two points in space. The first approach involves setting up and calculating the electric potential difference using the definition </p>
<table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]V(B) - V(A) = - \int _ A^ B \vec{\textbf{E}} \cdot d\vec{\textbf{s}} \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
In order to apply this result you must first calculate the electric field using the integral version of Coulomb's Law </p>
<table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\vec{E}\left(\vec{r}\right) = \frac{1}{4 \pi \epsilon _0} \int \frac{\rho (\vec{r}') \left(\vec{r} - \vec{r}'\right)}{|\vec{r} - \vec{r}'|^3} dV \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Alternatively, for sufficiently symmetric distributions, you could calculate [mathjaxinline]\vec{E}[/mathjaxinline] using Gauss's Law and then integrate to get the potential difference. </p>
<p>
The other way to find potential differences is to use the fact that for a single point-like charged object with charge [mathjaxinline]dq[/mathjaxinline] located at the origin, the electric potential difference between infinity and the field point [mathjaxinline]P[/mathjaxinline] located a distance [mathjaxinline]r[/mathjaxinline] from the charged object is given by </p>
<table id="a0000000004" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]V(r) - V(\infty ) = - \int _\infty ^ r \vec{E} \cdot d\vec{s} = \frac{1}{4 \pi \epsilon _0} \frac{dq}{r} \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Now let's generalize this to a continuous source charge distribution. </p>
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<img src="/assets/courseware/v1/9ab216988ceadc8761688fbc1c3eeedf/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_ps03_v02-fig026.jpg" width="400"/>
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<p>
Consider an infinitesimal charge element [mathjaxinline]dq = \rho (\vec{r}')dV[/mathjaxinline], where [mathjaxinline]\vec{r}'[/mathjaxinline] is the source vector from the origin to the infinitesimal charge element. Let the vector [mathjaxinline]\vec{r}[/mathjaxinline] denote the vector from the origin to the field point [mathjaxinline]P[/mathjaxinline]. Then [mathjaxinline]|\vec{r} - \vec{r}'|[/mathjaxinline] is the distance from the infinitesimal charge element to the field point [mathjaxinline]P[/mathjaxinline]. We set [mathjaxinline]V(\infty ) = 0[/mathjaxinline]. Then the potential difference between infinity and the field point [mathjaxinline]P[/mathjaxinline] due to just the charge [mathjaxinline]dq[/mathjaxinline] is </p>
<table id="a0000000005" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]dV\left(\vec{r}\right) = \frac{1}{4 \pi \epsilon _0} \frac{dq}{|\vec{r} - \vec{r}'|} = \frac{1}{4 \pi \epsilon _0} \frac{\rho (\vec{r}') dV}{|\vec{r} - \vec{r}'|} \, , \qquad V(\infty ) = 0 \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
We now apply the superposition principle and therefore integrate over the source charge distribution and find that </p>
<table id="a0000000006" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]V\left(\vec{r}\right) = \frac{1}{4 \pi \epsilon _0} \int _{\text {source}} \frac{\rho (\vec{r}') dV}{|\vec{r} - \vec{r}'|} \, , \qquad V(\infty ) = 0 \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Given an expression for the potential at any location [mathjaxinline]\vec{r}[/mathjaxinline], the difference between two points can be easily found by simply subtracting the potential at one point from the potential at the other one. </p>
<p>
One of the most important learning objectives is for you to determine which of these approaches to choose. For some problems both approaches are manageable but one may be easier than the other. You should now apply these ideas to the following problems. </p>
<p><b class="bfseries">(Part a)</b> A thin disk with outer radius [mathjaxinline]R_{2} =3 \, cm[/mathjaxinline] has a circular hole of radius [mathjaxinline]R_{1} = 1 \, cm[/mathjaxinline] in the middle. There is a uniform negative surface charge density of [mathjaxinline]\sigma =-10^{-7 \, } C/m^2[/mathjaxinline] on the disk. </p>
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<p>
What is the electric potential at the center of the hole? Assume zero potential at infinite distance from the center, and give your answer in volts ([mathjaxinline]V[/mathjaxinline]). [Reminder: [mathjaxinline]k=\dfrac {1}{4\pi \varepsilon _0}=9.0\times 10^{9} \, N \cdot m^2 / C^{2}[/mathjaxinline]] </p>
<p>
<p style="display:inline">[mathjaxinline]V =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> An electron starts at a location very far away along the axis of the disk (far enough to be considered at [mathjaxinline]z\approx \infty[/mathjaxinline]). It is initially moving along the axis towards the disk with a velocity [mathjaxinline]\vec{v}[/mathjaxinline] in the [mathjaxinline]-\hat{z}[/mathjaxinline] direction. The electron experiences no forces except repulsion by the charges on the disk and is observed to stop exactly at the center of the hole. What was the initial speed [mathjaxinline]|\vec{v}|[/mathjaxinline] of the electron? The mass of the electron is [mathjaxinline]m_ e = 9.11 \times 10^{-31} \, \mathrm{kg}[/mathjaxinline]. Give your answer in [mathjaxinline]m/s[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]v =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> A non-conducting disk of radius [mathjaxinline]R[/mathjaxinline] lies in the <i class="itshape">xy</i>-plane and carries a non-uniform charge density [mathjaxinline]\sigma =\sigma _{0} R/r[/mathjaxinline]. Determine the electric potential on the axis of the disk a distance <i class="itshape">z</i> from its center. Check the limit where [mathjaxinline]z \gg R[/mathjaxinline] to see that you recover the point-charge formula. </p>
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<img src="/assets/courseware/v1/111c629510dcd9cffd1a1e3a90e7b6df/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_ps03_v02-fig030.png" width="220"/>
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Hint: the following integral may be helpful. </p>
<table id="a0000000013" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\int \frac{dr}{\sqrt {r^{2} +z^{2} } } =\ln \left[ r+\sqrt {r^{2} +z^{2} } \right][/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
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For the symbolic check, write your answer using some or all of the following: [mathjaxinline]R[/mathjaxinline], [mathjaxinline]z[/mathjaxinline], sigma_0 for [mathjaxinline]\sigma _0[/mathjaxinline], pi for [mathjaxinline]\pi[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] for [mathjaxinline]k \equiv 1/(4 \pi \epsilon _0)[/mathjaxinline]. You will need to use the natural logarithm function ln() and the square root function sqrt(). </p>
<p>
<p style="display:inline">[mathjaxinline]V(z) =[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">W4PS4: Electric Potential Difference for a Uniformly CHarged Slab</h2>
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Electric Potential Difference for a Uniformly Charged Slab
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Consider a semi-infinite slab of charge with uniform positive charge density [mathjaxinline]\rho &gt; 0[/mathjaxinline] that extends from [mathjaxinline]x = -d[/mathjaxinline] to [mathjaxinline]x = d[/mathjaxinline]. Find the electric potential difference [mathjaxinline]V \left( B \right) - V \left( P \right)[/mathjaxinline] between the points [mathjaxinline]B[/mathjaxinline] and [mathjaxinline]P[/mathjaxinline] that lie on the [mathjaxinline]x[/mathjaxinline]-axis. The point [mathjaxinline]B[/mathjaxinline] lies outside the slab with [mathjaxinline]x_ B &gt; d[/mathjaxinline] and the point [mathjaxinline]P[/mathjaxinline] lies at the center of the slab with [mathjaxinline]x_ P = 0[/mathjaxinline]. </p>
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Write your answer using some or all of the following: d, x_B for [mathjaxinline]x_ B[/mathjaxinline], rho for [mathjaxinline]\rho[/mathjaxinline], epsilon_0 for [mathjaxinline]\varepsilon _0[/mathjaxinline]. <p style="display:inline">[mathjaxinline]V \left( B \right) - V \left( P \right) =[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_fps_fridayw3_5_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<h2 class="hd hd-2 unit-title">W4PS5: Partial Derivatives and the Gradient</h2>
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Partial Derivatives and the Gradient
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<p>
For one-dimensional functions of a single variable, e.g. [mathjaxinline]f(x)[/mathjaxinline], the derivative, [mathjaxinline]df/dx[/mathjaxinline] or [mathjaxinline]f'(x)[/mathjaxinline], tells you how the function changes for small changes in the independent variable <i class="itshape">x</i>. When a function depends on several variables, e.g. [mathjaxinline]f(x,y,z)[/mathjaxinline], there are several different derivatives, called partial derivatives, [mathjaxinline]\partial f/\partial x[/mathjaxinline], [mathjaxinline]\partial f/\partial y[/mathjaxinline], and [mathjaxinline]\partial f/\partial z[/mathjaxinline]. Both in concept and in practice the partial derivative is the same as the one-dimensional derivative, asking how the function changes as you change one of its independent variables (while holding the others fixed - treat them as constants when you take the derivative). </p>
<p>
Consider a point-like positively charged object with charge [mathjaxinline]+q[/mathjaxinline] located at the origin. Choosing [mathjaxinline]V(\infty )=0[/mathjaxinline], one can show that the electric potential difference [mathjaxinline]V(r)-V(\infty )=V(r)[/mathjaxinline] between infinity and any point on a sphere of radius [mathjaxinline]r[/mathjaxinline] centered on the origin is given by the expression </p>
<table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]V(r)=\frac{1}{4\pi \epsilon _0}\frac{q}{r} \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p><b class="bfseries">(Part a)</b> Find an expression for the potential in Cartesian coordinates [mathjaxinline](x,y,z)[/mathjaxinline]. </p>
<p>
For the symbolic check, write your answer using some or all of the following: [mathjaxinline]x[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], [mathjaxinline]z[/mathjaxinline], [mathjaxinline]q[/mathjaxinline] and k for [mathjaxinline]k=1/(4\pi \epsilon _0)[/mathjaxinline]. Use sqrt() as necessary. </p>
<p>
<p style="display:inline">[mathjaxinline]V(x,y,z) =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Calculate the three derivatives, called partial derivatives, [mathjaxinline]\partial V/\partial x[/mathjaxinline], [mathjaxinline]\partial V/\partial y[/mathjaxinline], and [mathjaxinline]\partial V/\partial z[/mathjaxinline]. </p>
<p>
For the symbolic check, write your answer using some or all of the following: [mathjaxinline]x[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], [mathjaxinline]z[/mathjaxinline], [mathjaxinline]r[/mathjaxinline], [mathjaxinline]q[/mathjaxinline] and k for [mathjaxinline]k=1/(4\pi \epsilon _0)[/mathjaxinline]. Use sqrt() as necessary. If the expression [mathjaxinline](x^2+y^2+z^2)^{\frac{1}{2}}[/mathjaxinline] appears, replace it with [mathjaxinline]r[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle { \frac{\partial V}{\partial x}} =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\displaystyle { \frac{\partial V}{\partial y}} =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\displaystyle { \frac{\partial V}{\partial z}} =[/mathjaxinline] </p>
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<p>
<b class="bfseries">Gradient</b>
</p>
<p>
It is convenient to define the gradient of a multidimensional scalar function. The gradient is a vector field that everywhere points in the direction of maximum change (steepest ascent uphill) of the function. It is calculated as follows: </p>
<table id="a0000000007" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\vec{\nabla } f=\frac{\partial f}{\partial x} \hat{\mathbf{i}} +\frac{\partial f}{\partial y} \hat{\mathbf{j}} +\frac{\partial f}{\partial z} \hat{\mathbf{k}}.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p><b class="bfseries">(Part c)</b> Again consider [mathjaxinline]V(x,y,z)=\dfrac {k q}{\sqrt {x^{2} +y^{2} +z^{2} } }[/mathjaxinline]. Calculate its gradient [mathjaxinline]\vec{\nabla } V[/mathjaxinline]. Use the fact that [mathjaxinline]\vec{\mathbf{r}} =x~ \hat{\mathbf{i}} +y~ \hat{\mathbf{j}} +z~ \hat{\mathbf{k}}[/mathjaxinline] and [mathjaxinline]r=\sqrt {x^{2} +y^{2} +z^{2} }[/mathjaxinline] to express your answer in terms of [mathjaxinline]k[/mathjaxinline], [mathjaxinline]q[/mathjaxinline], [mathjaxinline]\vec{\mathbf{r}}[/mathjaxinline], and [mathjaxinline]r[/mathjaxinline]. </p>
<p>
For the symbolic check, write your answer using some or all of the following: [mathjaxinline]r[/mathjaxinline], [mathjaxinline]q[/mathjaxinline], vecr for [mathjaxinline]\vec{\textbf{r}}[/mathjaxinline] and k for [mathjaxinline]k=1/(4\pi \epsilon _0)[/mathjaxinline]. Use sqrt() as necessary. </p>
<p>
<p style="display:inline">[mathjaxinline]\vec{\nabla } V =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part d)</b> Based on your answer to <b class="bfseries">(Part c)</b>, what is the relationship between [mathjaxinline]\vec{\nabla } V[/mathjaxinline] and the electric field [mathjaxinline]\vec{\mathbf{E}}[/mathjaxinline]? </p>
<p>
There is a solution but no answer checking for this part. </p>
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<p><b class="bfseries">(Part e)</b> (i) For a point-like charged object, does the gradient [mathjaxinline]\vec{\nabla } V[/mathjaxinline] point in the direction of maximum increase or maximum decrease of the function [mathjaxinline]V[/mathjaxinline]? Explain your reasoning. </p>
<p>
(ii) Does the sign of the charge effect your answer to part (i)? Explain your reasoning. </p>
<p>
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<p>
<b class="bfseries">Generalization: Finding the Electric Field from the Electric Potential</b>
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<p>
Your results for the point-like charged object generalize to any source of electric field. Because the electrostatic force on a charged object in an electrostatic field is a conservative force, the line integral [mathjaxinline]\int \nolimits _{A}^{B}\vec{\mathbf{F}} _{elec} \cdot d\vec{\mathbf{s}}[/mathjaxinline] is path independent and hence in class we defined the electric potential difference by </p>
<table id="a0000000009" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]V(B)-V(A)=-\int _{A}^{B}\frac{\vec{\mathbf{F}}}{q_{t} } \cdot d\vec{\mathbf{s}} = -\int _{A}^{B} \vec{\mathbf{E}} \cdot d\vec{\mathbf{s}} \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Whenever a line integral of a function is independent of the path, the line integral can be written as </p>
<table id="a0000000010" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]V(B)-V(A)=\int \nolimits _{A}^{B}\vec{\nabla } V\cdot d\vec{\mathbf{s}} \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Comparing our two expressions we see that </p>
<table id="a0000000011" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\vec{\textbf{E}} = - \vec{\nabla } V \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p><b class="bfseries">(Part f)</b> Using the result that [mathjaxinline]\vec{\textbf{E}} = - \vec{\nabla }V[/mathjaxinline], does the electric field depend on your choice of point for the zero reference electric potential? Explain your reasoning. </p>
<p>
There is a solution but no answer checking for this part. </p>
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<h2 class="hd hd-2 unit-title">W4PS6: Equipotentials</h2>
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Reading Equipotentials (Parts a and b)
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<p>The figure shows some equipotential lines around two charges, [mathjaxinline]q_1[/mathjaxinline] and
[mathjaxinline]q_2[/mathjaxinline]. Such lines give you a topographic map of the electric field. At places where lines
are closely spaced, i.e. where the potential gradient is steeper, the field is more intense than where the lines are
more widely spaced. The lines of equipotential of an electric field also tell you how much the potential energy of a
charge changes as it is moved from place to place in the field.</p>
<center>
<img src="/assets/courseware/v1/7e05fd276b8d49fc96fc047dc3e7d1aa/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_W04D1_2q_equipots.png" width="600"/>
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<p><b> (Part a) </b> What is the sign of the charge [mathjaxinline]q_1[/mathjaxinline]? </p>
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<p><b> (Part b) </b> Which charge is larger in magnitude?</p>
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(Parts c and d)
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<p><b> (Part c) </b> What is the change in potential energy of a charge [mathjaxinline]Q=-2.5\; \mu C[/mathjaxinline] when it is moved
from point C to B? (Give your answer in [mathjaxinline]\mu J[/mathjaxinline].) </p>
<p style="display:inline"> [mathjaxinline]\Delta U =[/mathjaxinline] </p>
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<p style="display:inline"> (in [mathjaxinline]\mu J[/mathjaxinline])</p>
<p><b> (Part d) </b> If a charge [mathjaxinline] Q=1.5 \, C [/mathjaxinline] is moved from point A to point C in the figure, how much work
does the field do? (Think about the sign and give your answer in [mathjaxinline] J [/mathjaxinline].) </p>
<p style="display:inline">[mathjaxinline]\Delta W =[/mathjaxinline] </p>
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<p style="display:inline"> (in [mathjaxinline] J [/mathjaxinline])</p>
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Grass Seeds
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Below we show the grass seeds representation of either the electric field or electric potential of two point charges. Which of the following statement(s) is correct? </p>
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<text> a) Both charges have the same sign, equal magnitude, and the grass seeds representation is the electric field.</text>
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<text> b) Both charges have the same sign, equal magnitude, and the grass seeds representation is the electric potential.</text>
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<input type="radio" name="input_exam1rev-cq5_2_1" id="input_exam1rev-cq5_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="exam1rev-cq5_2_1-choice_3-label" for="input_exam1rev-cq5_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_exam1rev-cq5_2_1">
<text> c) The charges have opposite signs, equal magnitude, and the grass seeds representation is the electric field.</text>
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<text> d) The charges have opposite signs, equal magnitude, and the grass seeds representation is the electric potential.</text>
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<text> e) Both charges have the same sign, unequal magnitude, and the grass seeds representation is the electric field.</text>
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<text> f) Both charges have the same sign, unequal magnitude, and the grass seeds representation is the electric potential.</text>
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<text> g) The charges have opposite signs, unequal magnitude, and the grass seeds representation is the electric field.</text>
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<text> h) The charges have opposite signs, unequal magnitude, and the grass seeds representation is the electric potential.</text>
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