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29 December, 06:07

The plates of a parallel-plate capacitor are 3.50 mm apart, and each carries a charge of magnitude 75.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 5.00*10^6 V/m. a. What is the potential difference between the plates? b. What is the area of each plate? c. What is the capacitance?

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  1. 29 December, 06:13
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    Vab = 17.5kV

    A = 16.9 cm2

    C = 4.27pF

    Explanation:

    a) Find the voltage difference:

    Vab = Ed

    E Electric field

    d distance between plates

    Vab potential difference

    d = 3.5mm

    = 3.5 * 10^ (-3) m

    Q = 75.0nC

    = 75 * 10^ (-9)

    E = 5.00 * 10^6 V/m

    Vab = (5.00 * 10^6) * (3.5 * 10^ (-3))

    = 17.5 * 10^3 V

    =17.5kV

    b. What is the area of the plate?

    The relation between the electric field and area is given as:

    E = Q / (ϵ0 * A)

    A = Q / (ϵ0 * E)

    Where ϵ0 is the electric constant and equals 8.854 * 10^ (-12) C2/N•m2

    A = 75 * 10^ (-9) / (8.854 * 10^ (-12) (5.00 * 10^6)

    = 1.69 X 10^ (-3) m2

    = 16.9 cm2

    c. Find the capacitance

    The equation relating capacitance, area of plate and plate distance is given by:

    C = ϵ0 A/d

    plug in the values of d, ϵ0 and A above to get the capacitance:

    C = (8.854 * 10^ (-12) * 1.69 X 10^ (-3) / 3.5 * 10^ (-3)

    = 4.27 * 10^ (-12) F

    = 4.27pF
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