A parallel plate capacitor above has square plates 10 cm across on each side. The plates are 50 μm apart, and the electric field strength in

Question

A parallel plate capacitor above has square plates 10 cm across on each side. The plates are 50 μm apart, and the electric field strength inside the capacitor is 4.0 x 10⁵ V/m.
A. What is the electric potential difference (voltage) across the capacitor?
B. What is the magnitude of the charge on either plate of the capacitor?
C. What is the capacitance of this capacitor?

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Nguyệt Ánh 4 years 2021-08-27T23:30:13+00:00 2 Answers 9 views 0

Answers ( )

  1. haidang
    0
    2021-08-27T23:31:58+00:00
    Reply

    Answer:

    a) ΔV = 20 V
    , b) Q = 35.4 10⁻⁹ C, c)  C = 1.77 10⁻⁹ F

    Explanation:

    a) The electric potential is

                     ΔV = E x

                     ΔV = 4.0 10⁵ 50 10⁻⁶

                      ΔV = 20 V

    c) The capacity of a capacitor is

             C = ε₀  A / d

    The area of ​​a square plate is

           A = L²

           A = 0.10²

           A = 0.01 m²

           L = 50mm = 50 10⁻⁶ m

    Let’s calculate

            C = 8.85 10⁻¹²  0.01 /50 10⁻⁶

            C = 1.77 10⁻⁹ F

    b)  the charge is

                 Q = C ΔV

                 Q = 1.77 10⁻⁹ 20

                  Q = 35.4 10⁻⁹ C

  2. Xavia
    0
    2021-08-27T23:32:03+00:00
    Reply

    Answer:

    A. V= 20 V  B. = 35.4 nC  C. =1.77 nF

    Explanation:

    A)

    • As the electric potential is the work done by the electric force through a distance d, per unit charge, and the force per unit charge, is the electric field, if we assume that the electric field is constant and perpendicular to the plates, we can express the potential difference between the plates of the capacitor as follows:

            V = E*d

    • where E is the magnitude of the electric field produced by the charge on one of the plates, and d, the separation between both plates.
    • Replacing by the givens, we have:

            V = E*d = 4e5 V/m* 50e-6 m = 20 V

    B)

    • Applying Gauss ‘s  Law to a gaussian surface with a shape of a pillbox, half outside the plates, half inside it , parallel to the surface, assuming that  the electric field is constant, and normal to the surface (outward directed from the positive charged plate), we can write the following expression, according to Gauss’Law:

           E*A = \frac{Q}{\epsilon_{0} }

    • Solving for Q:

           Q = E*A*\epsilon_{0} \\\\ Q = 4e5 N/C*1e-2m2*8.85e-12 C2/N*m2 = 35.4 nC\\

    C)

    • By definition, capacitance of any capacitor is as follows:

           C = \frac{Q}{V}

    • Replacing by the values found in A) and B) we have:

             C = \frac{Q}{V} = \frac{35.4 nC}{20 V} =1.77 nF

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