PHYSICS 2020 CAPA 5 - PHYS2020 Fall 2006 CAPA Set 5 1[1pt...

Info icon This preview shows pages 1–2. Sign up to view the full content.

PHYS2020, Fall 2006 CAPA Set 5 1. [1pt] The two plates of a capacitor hold +2710 μC and -2710 μC of charge, respectively, when the potential difference is 500 V . What is the capacitance? Given the charge and voltage, the capacitance can be calculated from C = Q/V . For these numbers, C = 2 . 71 × 10 - 3 C/ 500 V = 5 . 42 × 10 - 6 F . 2. [1pt] If a capacitor has 3.14 μC of charge on it and an electric field of 1.96 kV/mm is desired if they are separated by 3.53 mm of air, what must each plate’s area be? A constant electric field E (given) across a distance of d (also given) translates into a potential difference of V = Ed . Since we are also given the charge on the capacitor Q , we can find the required capacitance C = Q/V = Q/ ( Ed ). The capacitance of a parallel plate capacitor is given by the geometry and filling material. For an air filled capacitor it is C = Aepsilon1 0 /d . Putting these two equations together we find Aepsilon1 0 /d = Q/ ( Ed ). Solving for A we get A = Q/ ( epsilon1 0 E ). Note that the distance between the plates, d cancels out. One could also derive this more directly by noting that for a single plate, E = 2 πkσ = σ/ (2 epsilon1 0 ) where σ = Q/A and the electric field is multiplied by 2 for two plates giving E = Q/ ( epsilon1 0 A ). For these numbers, the answer is A = 3 . 14 × 10 - 6 / (8 . 85 × 10 - 12 C 2 /Nm 2 · 1 . 96 × 10 6 V/m ) = 0 . 181 m 2 . 3. [1pt] A cardiac defibrilator is used to shock a heart that is beating erratically. A capacitor in this device is charged to 5930 V and stores 220 J of energy. What is its capacitance?
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern