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--- 184_notes:examples:week7_cylindrical_capacitor [2017/10/06 17:10] – [Finding the Capacitance of a Cylindrical Capacitor] tallpaul
+++ 184_notes:examples:week7_cylindrical_capacitor [2018/06/19 15:37] – curdemma
@@ Line 1: / Line 1: @@
+[[184_notes:cap_in_cir|Return to Capacitors in Circuit]]
 =====Finding the Capacitance of a Cylindrical Capacitor=====
 Find the capacitance of a cylindrical capacitor. The structure of the capacitor is a cylindrical shell inside another cylindrical shell. The two shells become oppositely charged when the capacitor is connected to a power source. The length of the cylinders is $L$, and their radii are $a$ and $b$, with $a<b$.
@@ Line 12: / Line 14: @@
 ===Approximations & Assumptions===
   * The cylinders are much longer than they are far from one another, i.e., $L >> a, b$.
+  * Cylinders are uniformly charged.
 ===Representations===
@@ Line 17: / Line 20: @@
   * We represent the situation below.
-{{ 184_notes:7_cylindrical_cap.png?200 |Cylindrical Capacitor}}
+[{{ 184_notes:7_cylindrical_cap.png?200 |Cylindrical Capacitor}}]
 ====Solution====
-Let's start with node $A$. Incoming current is $I_1$, and outgoing current is $I_2$. How do we decide if $I_{A\rightarrow B}$ is incoming or outgoing? We need to bring it back to the Node Rule: $I_{in}=I_{out}$. Since $I_1=8 \text{ A}$ and $I_2=3 \text{ A}$, we need $I_{A\rightarrow B}$ to be outgoing to balance. To satisfy the Node Rule, we set
+In order to find capacitance, we need a charge $Q$ and a potential difference $\Delta V$. But the potential difference depends on how much charge is on the cylindrical plates. The idea here is to use the charge that is on the plates and use that charge to then solve for $\Delta V$. When we compute the capacitance, the charge on the plates should cancel out so that the capacitance only depends on the shape/size of the cylinders. For now, we say there is a charge $Q$ on the inner cylinder, and a charge $-Q$ on the outer cylinder. Since the cylinders are conductors (as they would be in any capacitor), the charge is uniformly distributed on the cylindrical shells.
-$$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$
+In order to arrive at potential difference, we will need to find the electric field. Remember from the [[184_notes:pc_potential#Relating_Electric_Potential_to_Electric_Field|notes on electric potential]] that in general we can write the expression for potential difference as $$\Delta V=V_f-V_i=-\int_{r_i}^{r_f} \vec{E}\bullet \text{d}\vec{r}$$
+We will end up integrating in the radial direction (a convenient choice, as this is how the electric field is directed!) from $a$ to $b$, which will give us the potential difference between the two cylinders.
+In order to find the electric field between the cylinders, we will use Gauss' Law. Below, we show a Gaussian surface that is cylindrical and fits inside the capacitor, with a radius $s$, which $a<s<b$. We also show vectors for area and electric field for the top and wall of the cylinder.
+[{{ 184_notes:7_cylindrical_cap_gauss.png?250 |Cylindrical Capacitor with Gaussian Surface}}]
+We have done a [[184_notes:examples:week5_flux_cylinder_line|similar problem]] involving Gauss's Law before. In order to find the electric flux in terms of the electric field, we write $$\Phi=\int\vec{E}\bullet \text{d}\vec{A}$$
+The integral is over the entire Gaussian surface, upon which $\vec{E}\bullet \text{d}\vec{A}$ takes on different values. On the top and bottom of the Gaussian surface, the electric field is directed radially, whereas the area-vectors point up and down, respectively (as shown above). Both cases yield $\vec{E}\bullet \text{d}\vec{A}=0$, so we can safely ignore this part of the integral. On the wall of the Gaussian surface, the electric field and the area-vectors are //both// directed radially, so their dot product simplifies to $$\vec{E}\bullet \text{d}\vec{A} = E(s)\text{d}A$$
-We do a similar analysis for node $B$. Incoming current is $I_{A\rightarrow B}$, and outgoing current is $I_3$. Since $I_{A\rightarrow B}=5 \text{ A}$ and $I_3=4 \text{ A}$, we need $I_{B\rightarrow D}$ to be outgoing to balance. To satisfy the Node Rule, we set
+We write $E(s)$ because the electric field magnitude depends only on the distance from the central vertical axis. Note, this notation is read as "electric field as a function of s" and NOT as "electric field times s". Another way to think about this is to say we can rotate the entire system about this axis and it does not change, so $E$ cannot depend on the angle $\phi$. We can also shift the entire system up or down along the axis, and it does not change (using the "$L$ is very large" assumption), so $E$ cannot depend on $z$. All that remains is $r$ or how far away you are from the cylinder. And since all along the Gaussian surface, $r=s$, we write $E(s)$.
-$$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$
-For node $C$, incoming current is $I_2$ and $I_3$. There is no outgoing current defined yet! $I_{C\rightarrow D}$ must be outgoing to balance. To satisfy the Node Rule, we set
+At this point, we can apply Gauss' Law. We equate flux to the charge enclosed divided by $\epsilon_0$:
-$$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$
+\begin{align*}
+\frac{Q_{enclosed}}{\epsilon_0} &= \int\vec{E}\bullet \text{d}\vec{A} \\
+                                 &= \int E(s)\text{d}A \\
+                                 &= E(s) \int \text{d}A \\
+                                 &= E(s) A_{wall} \\
+                                 &= E(s) (2\pi sh)
+\end{align*}
-Lastly, we look at node $D$. Incoming current is $I_{B\rightarrow D}$ and $I_{C\rightarrow D}$. Since there is no outgoing current defined yet, $I_{D\rightarrow battery}$ must be outgoing to balance. To satisfy the Node Rule, we set
+The last thing we need is $Q_{enclosed}$. This is simply the fraction of $Q$ that the Gaussian surface encloses. Since the height of the Gaussian cylinder is $h$, we have $Q_{enclosed}=\frac{h}{L}Q$. We can now write the magnitude of the electric field at a radius $s$ from the central vertical axis (given that $a<s<b$).
-$$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$
+$$E(s)=\frac{Q_{enclosed}}{\epsilon_0}\frac{1}{2\pi sh} = \frac{Q}{2\pi\epsilon_0 L s}$$
-Notice that $I_{D\rightarrow battery}=I_1$. This will always be the case for currents going in and out of the battery (approximating a few things that are usually safe to approximate, such as a steady current). In fact, we could have treated the battery as another node in this example. Notice also that if you incorrectly reason about the direction of a current (incoming or outgoing), the calculation will give a negative number for the current. The Node Rule is self-correcting. A final diagram with directions is shown below.
+This is the magnitude of the electric field between the conducting cylinders. We can now find the potential difference between the plates. Notice below that we simplify the dot product in the same way we simplified the flux through the wall of the Gaussian surface.
+\begin{align*}
+|\Delta V| &= \int_{r_i}^{r_f} \vec{E}\bullet \text{d}\vec{r} \\
+           &= \int_a^b E(s)\text{d}s \\
+           &= \frac{Q}{2\pi\epsilon_0 L} \int_a^b \frac{1}{s}\text{d}s \\
+           &= \left. \frac{Q}{2\pi\epsilon_0 L} \log (s)\right|_{s=a}^{s=b} \\
+           &= \frac{Q}{2\pi\epsilon_0 L}\log\left(\frac{b}{a}\right)
+\end{align*}
-{{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}}
+Finally, we can find the capacitance based on the charge of one cylinder and the potential we just found:
+$$C=\frac{Q}{\Delta V} = \frac{2\pi\epsilon_0 L}{\log\left(\frac{b}{a}\right)}$$
+Notice that the capacitance no longer depends on the charge Q, but only on the shape/size of our capacitor ($a$, $b$, and $L$ )