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--- 184_notes:examples:week2_conducting_insulating_balls [2017/08/24 23:03] – [Example: Attempting to Charge Insulators by Induction] tallpaul
+++ 184_notes:examples:week2_conducting_insulating_balls [2018/02/03 21:15] – tallpaul
@@ Line 1: / Line 1: @@
 ===== Example: Attempting to Charge Insulators by Induction =====
-In the notes on [[184_notes:charge_and_matter|Charges and Matter]], we saw how to charge a pair of conductors using induction. The relevant figure is copied to the right.
+In the notes on [[184_notes:charging_discharging#Charging|Charging and Discharging]], we saw how to charge a pair of conductors using induction. The relevant figure is shown below as a representation. Is it possible to charge a pair of insulators using induction? Why or why not?
-{{  184_notes:induction.png?300|Induction with Conductors}}
-Is it possible to charge a pair of insulators using induction? Why or why not?
 ===Facts===
-  * The Avogadro constant is $N_A = 6.022 \cdot 10^{23} \text{ mol}^{-1}$
+  * Electrons in an insulator are tightly bound to the nucleus, so the atoms can polarize but charges cannot move freely through an insulator.
-    * Note: When we write the unit as $\text{ mol}^{-1}$, we mean particles per mole. We could also write this unit as $mol^{-1}=\frac{1}{mol}$.
-  * All electrons have the same charge, which is $e = -1.602\cdot10^{-19} \text{ C}$.
-===Lacking===
-  * Total Charge
-===Approximations & Assumptions===
-  * None, we have all the information we need.
 ===Representations===
-  * The total number of particles $N$ can be found from the number of moles $m$ using the Avogadro constant: $N = m \cdot N_A$.
+  * From the notes, we can pull a representation for how we would charge //conductors// using induction:
-  * The total charge $Q$ can be written as the number of particles $N$ times the charge of each particle ($e$, for electrons): $Q=N\cdot e$.
+{{ 184_notes:induction.png?300 |Induction with Conductors}}
+  * We can model the atoms in an insulator as little ovals (like the one below), that show when one side of the atom is more positive or negative than the other side. When ovals are not shown, this will just mean the atoms are not polarized.
+{{ 184_notes:polarizedatom.png?100 }}
+===Goal===
+  * Create an explanation for whether it is possible to charge a pair of insulators using induction.
 ====Solution====
-yeas
+<WRAP TIP>
+=== Assumption ===
+Right away, we make the assumption that an induction process for insulators would look the same as it does for conductors. The reason we make this assumption is because this process, as described in the notes, is our basis for understanding how charging by induction works. If it looks the same and involves the same steps, we can more easily describe what we think will happen.
+</WRAP>
+We show the analogous "induction with insulators" with a step-by-step representation below. We knew from the facts that electrons cannot move freely between insulators, which is one of the key differences between insulators and conductors. At time $t=t_0$, both of the insulating balls start out as neutral. At time $t=t_1$, when the connected balls are moved close the charged object, the atoms in the insulators would polarize, but the electrons are not free to move further or to move from one ball to the next. This means when the insulating balls are separated at $t=t_2$, there are two polarized but overall neutral balls. As the balls are pulled farther away from the positive charge, they become less and less polarized, eventually returning to the same state that they were before at $t=t_0$ (neutral, not polarized).
+{{ 184_notes:inducing_insulators.png?300 |Induction with Insulators}}
+The critical difference between conductors and insulators is that electrons can flow from one conductor to the other, but for insulators the electrons are bound to their nuclei. Because of this, the insulators do not charge by induction.