weber173_Lab4

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Dec 6, 2023

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Physics 272 – Lab 04 Charge Induction and Coulomb’s Law Can’t provide photograph of experimental setting/set up because I couldn’t conduct the lab this week! Q1: Because a static charge was created on the Styrofoam plate when you are placed it in front of the aluminum pan, an electrostatic force is present between them. Describe what is happening in steps A to F regards to the charges and distance. Explain in which case A through F, the force between the plate and the pan, is highest. From A-C, the distance between the plates was decreasing, so the magnitude of the force increased. At C, the pie pan is quite close to the Styrofoam, and the pan is discharged after grounding as the spark occurs. The charge increases as the aluminum pan goes from the C-E, and the magnitude of the force becomes smaller as distance increases. At F, a spark occurs as the pie pan is discharged again, so the force between Page 1 of 2
the two becomes approximately zero. The force is highest at D, as the distance between the two materials is the smallest. Q2: Assume that in step C the charge on the plate and the pan are equal and opposite (i.e., since the separation is small compared to the radius of the plates); therefore, these will behave like a charge sheet next to a grounded conducting plane. Use the Coulomb force between the plate and the pan to find the magnitude of the charge Q on each after step C. Since the separation is small as compared to the radius of the plate, we can assume E=Q a2ε 0 . By combining the equation F = EQ with the assumed equation, we can find the charge of Q. If we assume the radius of the plate to be 0.07m and F = 0.055N after step C (at step D), then Q = 0.00000012244. The pie pan would have an equal and opposite charge because the Styrofoam plate would have a negative charge after being rubbed by the cloth. Q3: After Step E the pan and plate are separated by a big distance compared to their sizes, so let's assume these will behave like point charges and the force between them will be about F E = Q 2 4 π ε 0 1 R 2 . Use the Q calculated in Q2: a separation of about 30 cm (1 foot), to calculate the electrostatic force. Compare your calculation with the value you got using the iOLab application graphs. By using the Q calculated in Q2, which we found to be Q = 1.2244E-7. When the R = 0.3m, F = - 0.0015 N. At step E on the graphs, the value of the force is very close to zero, so this calculated value is not exact, but there isn’t much of a difference. Q4: After step F, the missing charge on the pan is restored by the spark (i.e., a brief grounding). Since any charge separation within the pan is now small compared to the separation of the pan and the plate, how will the force between them change? Explain your responses? Since the charge separation within the pan is small, the force becomes approximately zero. The first spark caused the magnitude of the force to become larger, so when charge is restored by the second spark, the magnitude of the force becomes smaller. Page 2 of 2
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