a.What electric potential difference exists across a 5.2 µF capacitor that has a charge of 2.1 10-3 C?
how do i use uF and C into a formula to find answer??

b.An oil drop is negatively charged and weighs 8.5 10-15 N. The drop is suspended in an electric field intensity of 5.3 102 N/C.
(a) What is the charge on the drop?
(b) How many electrons does it carry?

c.The electric field intensity between two large, charged, parallel metal plates is 7500 N/C. The plates are 0.05 m apart. What is the electric potential difference between them?

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  1. a. Q = C V
    Therefore V = Q/C
    Q is the charge in coulombs
    C is the capacitance in farads, which in your case is 5.2*10^-6
    V will be in volts

    b. m g = Q E
    E = 530 n/C
    m g = 8.5810^-15 N
    Q = n e where e is the electron charge, 1.602*10^-19 C
    Solve for n.
    The number of electrons, n, that you get should be a small integer, or nearly so.

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  2. C = q/V
    that is definition of capacitor, charge in coulombs over voltage in volts

    5.2*10^-6 farads = 2.1 *10^-3 coulombs / v
    V = 2.1 * 10^-3 / 5.2 *10^-6
    = .404 * 10^3
    = 404 volts

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  3. Kailey - I showed you how to do part c an hour ago.

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