Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt poten

Question

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1000 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 60 volts

in progress 0
Thành Công 1 year 2021-09-03T04:21:35+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-09-03T04:23:28+00:00

    Answer:

    [tex]1.62\times 10^{-8}\ \text{s}[/tex]

    Explanation:

    [tex]\epsilon_0[/tex] = Vacuum permittivity = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]

    [tex]A[/tex] = Area = [tex]10\times 2\times 10^{-4}\ \text{m}^2[/tex]

    [tex]d[/tex] = Distance between plates = 1 mm

    [tex]V_c[/tex] = Changed voltage = 60 V

    [tex]V[/tex] = Initial voltage = 100 V

    [tex]R[/tex] = Resistance = [tex]1000\ \Omega[/tex]

    Capacitance is given by

    [tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}[/tex]

    We have the relation

    [tex]V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}[/tex]

    The time taken for the potential difference to reach the required level is [tex]1.62\times 10^{-8}\ \text{s}[/tex].

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )