Lab 1 A Law of Electrostatic Force

Rishi Kothari Physics 136-2 Lab TA: Vishnoi Lab Partner: Max Rothfeder, Ryan Boyle Page | 3 of 𝑘 1 . (Note: in a future lab we will talk about uncertainty in your slope, but for now you do not need to record that. Just record the slope itself.) You now know how the force depends on ? 1 . In principle, you should also verify how the force depends on ? 2 . But intuition suggests that “electric charge is electric charge.” Why should the electrostatic force depend differently on ? 2 than on ? 1 ? This implies that the force depends linearly on both charges, as: 𝐹 = 𝑘 12 ? 1 ? 2 (2) [Side note: of course, it is possible that the F vs. ? 1 graph is not a perfectly straight line after all, and that the relation between the force and charges is entirely different. This is particularly likely if our range of charges covers only a small part of what is possible in nature, so that our linear fit is only an approximation. For example, we didn’t test values larger than 9 μC or closer to zero than 3 µC, so it’s possible that something strange happens there. Ideally, we would verify the linear dependence with additional data, but you do not need to do this now.] Slope: in units of 33.703 N/μC \f

Rishi Kothari Physics 136-2 Lab TA: Vishnoi Lab Partner: Max Rothfeder, Ryan Boyle Page | 6 Results The electrostatic force law is known to be 𝐹 = 1 4πϵ 0 ? 1 ? 2 ? 2 (5) ϵ 0 is a fundamental physical constant, called the “vacuum permittivity” or the “permittivity of free space.” Other scientists have carefully measured ϵ 0 to have a value of 8.8542 × 10 −12 ? 2 /𝑁𝑚 2 . We have now collected enough data to determine this natural constant for ourselves. 3.1 (10 pts) In the first part of this lab, you determined 𝑘 1 , defined by 𝐹 = 𝑘 1 ? 1 . Thus 𝑘 1 = 1 4πϵ 0 ? 2 ? 2 (6) with ? 2 and ? having the fixed values you selected in 1.1 above. Use the value of 𝑘 1 you wrote down in 1.3 above to calculate ϵ 0 and enter it here. Please express ϵ 0 in units of ? 2 /𝑁𝑚 2 , which may require converting from cm or μC. 3.2 (10 pts) Comparing Equation (5) with our expressions 𝐹 = 𝑘 12 ? 1 ? 2 and 𝑘 12 = ?? 𝐵 from Equations (2) and (3), we get ? = 1 4πϵ 0 (7) Use the value of ? you entered in 2.3 to calculate ϵ 0 and enter it here. Again, please express ϵ 0 in units of ? 2 /𝑁𝑚 2 , which may require converting from cm or μC. 6cm=0.06m K1=33.703 N/μC = 33703000 N/C ∈= 8.854 × 10 −12 ? 2 /𝑁𝑚 2 ∈= 10 −12 × 10 4 4𝜋 × 89.837 ∈= 8.858 × 10 −12 ? 2 /𝑁𝑚 2