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10 December, 15:09

he 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 do

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  1. 10 December, 18:45
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    Incomplete question

    This is the complete question

    The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 V potential difference is suddenly applied to the initially uncharged plates through a 1000 Ω resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95 V?

    Explanation:

    Given that,

    The dimension of 10cm by 2cm

    0.1m by 0.02m

    Then, the area is Lenght * breadth

    Area=0.1*0.02=0.002m²

    The distance between the plate is d=1mm=0.001m

    Then,

    The capacitance of a capacitor is given as

    C=εoA/d

    Where

    εo is constant and has a value of

    εo = 8.854 * 10-12 C²/Nm²

    C = 8.854E-12*0.002/0.001

    C=17.7*10^-12

    C=17.7 pF

    Value of resistor resistance is 1000ohms

    Voltage applied is V = 100V

    This Is a series resistor and capacitor (RC) circuit

    In an RC circuit, voltage is given as

    Charging system

    V=Vo[1 - exp (-t/RC) ]

    At, t=0, V=100V

    Therefore, Vo=100V

    We want to know the time, the voltage will deflect 95V.

    Then applying our parameters

    V=Vo[1 - exp (-t/RC) ]

    95=100[1-exp (-t/1000*17.7*10^-12) ]

    95/100=1-exp (-t/17.7*10^-9)

    0.95=1-exp (-t/17.7*10^-9)

    0.95 - 1 = - exp (-t/17.7*10^-9)

    -0.05=-exp (-t/17.7*10^-9)

    Divide both side by - 1

    0.05=exp (-t/17.7*10^-9)

    Take In of both sides

    In (0.05) = -t/17.7*10^-9

    -2.996=-t/17.7*10^-9

    -2.996*17.7*10^-9=-t

    -t=-53.02*10^-9

    Divide both side by - 1

    t = 53.02*10^-9s

    t=53.02 ns

    The time to deflect 95V is 53.02nanoseconds
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