Differences

This shows you the differences between two versions of the page.

--- 184_notes:examples:week8_cap_series [2017/10/11 13:53] – [Example: Resistors in Series] tallpaul
+++ 184_notes:examples:week8_cap_series [2021/06/21 18:22] – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:c_series|Return to capacitors in series notes]]
 =====Example: Capacitors in Series=====
-Suppose you have the following circuit. Capacitors are labeled 1 and 2 for convenience of reference. You know that the circuit contains a 12-Volt battery, $Q_1=4.5 \mu\text{C}$, and $C_2=0.5 \mu\text{F}$. What is the capacitance of Capacitor 1? What happens to the charge on the capacitors if we insert a dielectric material with dielectric constant $k = 3$ into Capacitor 1?
+Suppose you have the following circuit. Capacitors are labeled 1 and 2 for convenience of reference. You know that the circuit contains a 12-Volt battery, $Q_1=4.5 \mu\text{C}$, and $C_2=0.5 \mu\text{F}$. What is the capacitance of Capacitor 1? What happens to the charge on //both// of the capacitors if we insert a dielectric material with dielectric constant $k = 3$ into Capacitor 1?
-{{ 184_notes:8_cap_series.png?300 |Circuit with Capacitors in Series}}
+[{{ 184_notes:8_cap_series.png?300 |Circuit with Capacitors in Series}}]
 ===Facts===
@@ Line 18: / Line 20: @@
   * We are measuring things like potential differences and charges when the circuit is in a steady state.
   * Approximating the battery as a mechanical battery.
+  * Capacitors are parallel plate capacitors.
 ===Representations===
@@ Line 26: / Line 29: @@
 ====Solution====
+===Part 1 ===
 Let's find $C_1$. In order to use the equation for equivalent capacitance of capacitors in series, as we have here, we first need the equivalent capacitance of the entire circuit. Remember that the [[184_notes:c_series#Node_Rule_and_Charge_in_Series|charge on the capacitors]] in series should be the same, so $Q_{\text{equiv}}=Q_1=Q_2$. Now, we can write:
 $$C_{\text{equiv}}=\frac{Q_{\text{equiv}}}{\Delta V_{\text{bat}}}=0.375 \mu\text{F}$$
-Now we can solve for C_2 using $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}$$
+Now we can solve for $C_1$ using $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}$$
-This gives us $C_1 = 0.5 \mu\text{F}$
+This gives us $C_1 = 1.5 \mu\text{F}$
-{{ 184_notes:8_cap_series_dielectric.png?300 |Circuit with Capacitors in Series}}
+===Part 2===
+[{{  184_notes:8_cap_series_dielectric.png?300|Circuit with Capacitors in Series, one capacitor has a dielectric}}]
-Now we need to consider what happens when we insert a dielectric. It might look something like the circuit above. A description of what a dielectric does in a capacitor is [[184_notes:cap_charging#Dielectrics|here]]. Its [[184_notes:cap_in_cir#Finding_Capacitance|effect on capacitance]] is: $$C =\frac{k\epsilon_0 A}{d}$$
+Now we need to consider what happens when we insert a dielectric. It might look something like the circuit to the right. A description of what a dielectric does in a capacitor is [[184_notes:cap_charging#Dielectrics|here]]. Its [[184_notes:cap_in_cir#Finding_Capacitance|effect on capacitance]] is: $$C =\frac{k\epsilon_0 A}{d}$$
 So when we insert the dielectric, we have a new capacitance for Capacitor 1: $C_{\text{1, new}}=kC_1=4.5 \mu\text{F}$. To find the new value that the capacitors are charged to, we return to the equivalent capacitance of the circuit:
@@ Line 40: / Line 45: @@
 This yields $C_{\text{equiv, new}}=0.45 \mu\text{F}$. Now, we can find the new charge:
 $$Q_{\text{new}} = C_{\text{equiv, new}}\Delta V_{\text{bat}} = 5.4 \mu\text{C}$$
+**This is the charge on __both__ capacitors** since the capacitors are in series. So even if we insert a dielectric in only one of the capacitors, the charge on both will increase.