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184_notes:examples:week2_conducting_insulating_balls [2017/08/24 22:57] – tallpaul | 184_notes:examples:week2_conducting_insulating_balls [2017/08/28 18:33] – dmcpadden | ||
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- | ===== Example: | + | ===== Example: |
- | Charged conducting balls vs insulating balls | + | In the notes on [[184_notes: |
+ | {{ 184_notes: | ||
+ | Is it possible to charge a pair of insulators using induction? Why or why not? | ||
===Facts=== | ===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=== | ===Lacking=== | ||
- | * Total Charge | + | * An explanation for whether it is possible to charge a pair of insulators using induction. |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * None, we have all the information | + | * We will use the same induction process as we did for conductors. |
+ | * The insulators start out neutral, meaning there are no excess electrons on the surface or any unbounded electrons (all electrons have a corresponding positive nuclei). | ||
===Representations=== | ===Representations=== | ||
- | | + | {{184_notes: |
- | * The total charge $Q$ can be written | + | |
+ | * We can use a similar diagram | ||
====Solution==== | ====Solution==== | ||
- | yeas | + | {{ 184_notes: |
+ | We show the analogous " | ||
+ | |||
+ | 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. Therefore, the insulators do not charge by induction. |