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| 184_notes:fab_physics [2019/01/11 04:32] – pwirving | 184_notes:fab_physics [2019/01/11 04:41] – pwirving |
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| $$.5(m)(v)^2 = k \frac{q_{1}*q_{2}}{r_{sep}}$$ | $$.5(m)(v)^2 = k \frac{q_{1}*q_{2}}{r_{sep}}$$ |
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| $$.5(6.645*10^{-27})(1.28*10^{7})^2 = 8.99*10^9 \frac{150*2*(1.6*10^{-19}}){r_{sep}}$$ | $$.5(6.645*10^{-27})(1.28*10^{7})^2 = 8.99*10^9 \frac{-150*2*(-1.6*10^{-19})}{r_{sep}}$$ |
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| $$r_{sep}=-7.9331e+5 \quad {meters}$$ | $$r_{sep}=-7.9331e+5 \quad {meters}$$ |
| <WRAP tip> | <WRAP tip> |
| === Discussion Prompts === | === Discussion Prompts === |
| * **Question:** What is your system when you're calculating the speed of the cloud? Why did it change when you were looking for the shortest separation distance between the clouds? | * **Question:** What is your system when you were looking for the shortest separation distance between the clouds? |
| * **Answer:** Initially we only cared about the small cloud speeding up (so it was the only thing in the system), but when we were looking for the shortest separation distance, we needed to consider the interaction between the two clouds so both were in the system. | * **Answer:** We needed to consider the interaction between the two clouds so both were in the system. |
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| * **Question**: In this problem you used both electric potential energy and electric potential. How are those two ideas related? | * **Question**: In this problem, you used electric potential energy, what is meant by this? |
| * **Answer**: Electric potential energy is related to potential by a charge. Electric potential energy tells you about the interaction between the two charges. Electric potential tells you about how much energy there would be at a location (from a source charge) without knowing or having another charge to interact with. | * **Answer**: Electric potential energy tells you about the interaction between the two charges. |
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| * **Question:** Why would make sense to ask for the electric potential (or electric field) on top of the headquarters but not the electric potential energy (or electric force)? | * **Question**: What would you expect to happen if the balloon was not stabilized on the moon? Which would move faster? Which charge exerts a larger force (the large or small charge)? |
| * **Answer**: There is not a charge on top of the HQ so there would be nothing for the electric force/energy to act on (no interaction). Potential and field | * **Answer**: The smaller charge would also exert a force on the bigger charge and push it away. The small charge would still move faster than the big charge because it has a smaller mass - they experience the same force. |
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| * **Question:** How do $\vec{F}$, $\vec{E}$, $V$, and $U$ relate to each other? | |
| * **Answer:** Important points to notice/emphasize out of the rectangle: 1) force and energy are interactions between charges (require at least two charges!), 2) potential and field are only about the source charge (only one charge needed!) and the observation location, 3) force and field are vectors - good for if you care about direction, 4) energy and potential are scalars - mathematically easier to handle. | |
| {{:184_notes:evfugeneralrelation.png?200|}} | |
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| * **Question:** Is there anything that still doesn't make sense? Any lingering questions? | |
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| === Evaluation Questions === | |
| * **Question**: What would you have to change about the situation/problem to make it safe for the equipment? | |
| * **Answer**: You'd have to change the height or charge of the clouds to make it safe. | |
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| * **Question**: What would you expect to happen if the mountains were not there? Which would move faster? Which cloud exerts a larger force (the large or small cloud)? | |
| * **Answer**: The smaller cloud would also exert a force on the bigger cloud and push it away. The small cloud would still move faster than the big cloud because it has a smaller mass - they experience the same force. | |
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| === Extension Questions === | === Extension Questions === |
| * **Question**: How much force would the mountains have to exert on the big cloud to keep in place? | * **Question**: How much force would the moon have to exert on the big charge to keep it in place? |
| * **Answer**: Here would you have to use net force where the magnitude of the contact (or normal) force from the mountains should be equal to the magnitude of the electric force $F=\frac{k*q*Q}{r^2}$. This can tie into the next question about force diagrams | * **Answer**: Here would you have to use net force where the magnitude of the contact (or normal) force from the moon should be equal to the magnitude of the electric force $F=\frac{k*q*Q}{r^2}$. This can tie into the next question about force diagrams |
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| * **Question:** What does the free body (force) diagram look like for the large and small cloud? | * **Question:** What does the free body (force) diagram look like for the large and small charge? |
| * **Answer:** (They may have these diagrams from their process already.) From the diagrams it should be clear why the big cloud does not move/change speeds (there are two balancing forces - no change in momentum) but the small cloud does (only one force - so the small cloud has an acceleration or a changing momentum). | |
| {{:184_solutions:3a_largecloudforce.jpg?200|}} | |
| {{:184_solutions:3a_smallcloudforce.jpg?200|}} | |
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| * **Question:** What would you expect to happen to the charges in the ground when these two clouds are nearby? (What kind of charges exist in the ground?) | |
| * **Answer**: The ground is made up of a lot of positive and negative charges. When there is a large negative charge above the ground, we would expect the positive charges in the ground to be attracted to the surface and the negative charges to be repelled away (polarizing). Depending on water content, the ground may act like an insulator or a conductor so the degree to which this happens varies. | |
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| </WRAP> | </WRAP> |