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184_notes:examples:week5_gauss_ball [2017/09/20 12:13] – [Solution (Part A)] tallpaul | 184_notes:examples:week5_gauss_ball [2021/06/07 13:54] – schram45 | ||
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+ | [[184_notes: | ||
+ | |||
=====Example: | =====Example: | ||
- | Suppose you charge | + | Suppose you have an insulating ball that has been charged somehow. The charging was very thorough and it seems like the ball is pretty much uniformly charged. It has a charge $Q$ and a radius $R$. What is the electric field at a distance $r$ from the center of the ball? Be sure to account for both cases $r< |
===Facts=== | ===Facts=== | ||
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* The " | * The " | ||
* $\vec{E}(\vec{r})$ | * $\vec{E}(\vec{r})$ | ||
- | |||
- | ===Approximations & Assumptions=== | ||
- | * The ball is a perfect sphere. | ||
- | * There are no other charges that affect our calculations. | ||
- | * The ball is not discharging. | ||
- | * The ball is uniformly charged throughout its volume. | ||
- | * For the sake of representation, | ||
===Representations=== | ===Representations=== | ||
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$$\Phi_e=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | $$\Phi_e=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | ||
* We represent the situation with the following diagram. The insulating case (part A) is shown on the left, and the conductor (part B) is on the right. | * We represent the situation with the following diagram. The insulating case (part A) is shown on the left, and the conductor (part B) is on the right. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
+ | |||
+ | <WRAP TIP> | ||
+ | ===Assumptions=== | ||
+ | * There are no other charges that affect our calculations. | ||
+ | * The ball is not discharging. | ||
+ | * For the sake of representation, | ||
+ | </ | ||
====Solution (Part A)==== | ====Solution (Part A)==== | ||
A step-by-step approach to using Gauss' Law is shown in the [[184_notes: | A step-by-step approach to using Gauss' Law is shown in the [[184_notes: | ||
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$$\vec{E}(\vec{r})=E(r)\hat{r}$$ | $$\vec{E}(\vec{r})=E(r)\hat{r}$$ | ||
Note that $r$ here is same $r$ in the premise of the problem: the distance from the center of the sphere. At this point, it becomes a little more clear what type of Gaussian surface we are after. If we choose the Gaussian surface to be a spherical shell centered with the ball, then its area-vectors will be parallel to $\vec{E}$, and $E(r)$ will be constant on its surface, since the shell' | Note that $r$ here is same $r$ in the premise of the problem: the distance from the center of the sphere. At this point, it becomes a little more clear what type of Gaussian surface we are after. If we choose the Gaussian surface to be a spherical shell centered with the ball, then its area-vectors will be parallel to $\vec{E}$, and $E(r)$ will be constant on its surface, since the shell' | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
- | The next step after choosing our Gaussian surface is to write the flux through the surface. Based on our symmetry argument above, the calculation should | + | The next step after choosing our Gaussian surface is to write the flux through the surface. Based on our symmetry argument above, the calculation should |
\begin{align*} | \begin{align*} | ||
\Phi_{\text{total}} &= \int\vec{E}\bullet \text{d}\vec{A} \\ | \Phi_{\text{total}} &= \int\vec{E}\bullet \text{d}\vec{A} \\ | ||
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&= E(r)\cdot 4\pi r | &= E(r)\cdot 4\pi r | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | <WRAP TIP> | ||
+ | ===Approximation=== | ||
+ | In order to take the electric field term out of the integral we must approximate our ball as a perfect sphere. Having a perfect sphere ensures all the area vectors are parallel to the electric field vectors. This also ensures that the electric field is constant through our Gaussian surface at a given radius. | ||
+ | </ | ||
+ | |||
Notice that the above result is the same for both $r<R$ and $r>R$. Next, we need to find the total charge enclosed by the Gaussian surface. Here, it is helpful to determine the charge density in our insulator. There is helpful section in the notes on [[184_notes: | Notice that the above result is the same for both $r<R$ and $r>R$. Next, we need to find the total charge enclosed by the Gaussian surface. Here, it is helpful to determine the charge density in our insulator. There is helpful section in the notes on [[184_notes: | ||
$$\rho=\frac{Q_{\text{total}}}{V_{\text{total}}}=\frac{Q}{\frac{4}{3}\pi R^3}=\frac{3Q}{4\pi R^3}$$ | $$\rho=\frac{Q_{\text{total}}}{V_{\text{total}}}=\frac{Q}{\frac{4}{3}\pi R^3}=\frac{3Q}{4\pi R^3}$$ | ||
+ | |||
+ | <WRAP TIP> | ||
+ | ===Assumption=== | ||
+ | Having a constant charge density is only possible if we assume that the ball is evenly charged throughout its volume. This may or may not be the case depending on how our insulator is charged. | ||
+ | </ | ||
+ | |||
The charge enclosed by the Gaussian surface, then, would relate to the charge density. For $r<R$, the surface does not enclose the entire ball, so we would just multiply the volume that the Gaussian surface encloses by the charge density to find the charge contained in the Gaussian surface. | The charge enclosed by the Gaussian surface, then, would relate to the charge density. For $r<R$, the surface does not enclose the entire ball, so we would just multiply the volume that the Gaussian surface encloses by the charge density to find the charge contained in the Gaussian surface. | ||
$$Q_{\text{enclosed}}(r< | $$Q_{\text{enclosed}}(r< | ||
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\end{cases} | \end{cases} | ||
\] | \] | ||
- | + | We are now fully equipped | |
- | First, we evaluate the situation qualitatively. Consider the electric field lines of the charged line: | + | $$E(r) = \frac{\Phi_{\text{total}}}{4\pi r^2} = \frac{Q_{\text{enclosed}}}{4\pi\epsilon_0 r^2}=\frac{Qr}{4\pi\epsilon_0 R^3}$$ |
- | It's a little tough to demonstrate the field lines with only two dimensions to draw on, but you can imagine that the thicker arrows point out of the page more, and the thinner arrows point into the page more. In essence, each arrow should point directly away from and perpendicular to the line of charge, as indicated in the formula | + | For $r>R$: |
- | + | $$E(r) | |
- | The electric field is parallel to the bases of the cylinder, so there are no electric field lines entering or exiting from the bases. So the flux through the bases should be $0$. For the wall of the cylinder, the electric field lines are perpendicular to the surface, which means they are parallel to the area-vectors. These facts will greatly simplify our integral calculation of the flux. | + | Together: |
- | \begin{align*} | + | \[ |
- | \Phi_{\text{total}} | + | E(r) = \begin{cases} |
- | &= 0 + \int_{\text{wall}}\vec{E}\bullet \text{d}\vec{A} \\ | + | \frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3} & r< |
- | &= \int_{\text{wall}}E\hat{r}\bullet \text{d}A\hat{r} \\ | + | |
- | & | + | |
- | &= EA_{\text{wall}} \\ | + | \] |
- | & | + | Outside the ball, the electric field exists as if the ball were a point charge! |
- | & | + | ====Solution (Part B)==== |
- | \end{align*} | + | We repeat the process above for the case that the ball is a conductor. Notice that much of the reasoning is the exact same. We still have spherical symmetry, and we choose the same Gaussian surface. It is pictured below for both $r<R$ and $r>R$. |
- | When we plug in values for $\lambda$, $l$, and $\epsilon_0$, we get $\Phi_{\text{cylinder}}=5.08\cdot 10^6\text{ Vm}$. | + | [{{ 184_notes:5_conductor_gaussian_surface.png? |
- | + | The only thing that is different here is the charge enclosed by the Gaussian surface. Since all the excess charge on a conductor hangs out on the surface, the inside of the conductor is neutral. So if $r<R$, then the charge enclosed by the Gaussian surface is $0$. If $r>R$, then the Gaussian surface encloses the whole ball, and the charge | |
- | Notice that in the [[184_notes:gauss_ex|next section of notes]], we define " | + | \[ |
- | $$\Phi_{\text{cylinder}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\lambda l}{\epsilon_0}$$ | + | Q_{\text{enclosed}} = \begin{cases} |
- | This is the same result! An alternative question for this example could have been: What is the electric field due to a uniformly charged line? If we were not given the electric field at the beginning, we could have used symmetry arguments and Gauss' Law to work backwards, starting with the charge enclosed, and then using the integral formula for electric flux to solve for the electric field. | + | 0 & r< |
+ | Q & r>R | ||
+ | \end{cases} | ||
+ | \] | ||
+ | We are now fully equipped to solve for the electric field. Remember that Gauss' Law states $\Phi_e=Q_{\text{enclosed}}/ | ||
+ | $$E(r) = \frac{\Phi_{\text{total}}}{4\pi r^2} = \frac{Q_{\text{enclosed}}}{4\pi\epsilon_0 | ||
+ | For $r>R$: | ||
+ | $$E(r) | ||
+ | Together: | ||
+ | \[ | ||
+ | E(r) = \begin{cases} | ||
+ | 0 & r< | ||
+ | | ||
+ | | ||
+ | \] | ||
+ | Outside |