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 975 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 75 volts

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Cherry 5 years 2021-08-04T13:03:29+00:00 2 Answers 45 views 0

Answers ( )

  1. thachthao
    0
    2021-08-04T13:04:32+00:00
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    Answer:

    t = 24.3ns

    Explanation:

    The deflection plates make up a parallel plate capacitor fed via 975 ohm resistor from a 100V supply

    capacitance C = (e0 x A) / d

    C = \frac{(8.85 X 10^{-12} X (2 X 10^{-2} X 10 X 10^{-2} ) )}{10^{-3} }

    C = C = 18 X 10^{-12} F

    Vc = Vx (1-e^{-(\frac{t}{CR} )} )

    75 = 100 X (1-e^{-(\frac{t}{CR} )} )

    \frac{75}{100} = (1-e^-{(\frac{t}{CR} )} )

    0.75 = (1-e^{-(\frac{t}{CR} )} )

    0.75 - 1 = -e^{-(\frac{t}{CR} )}

    e^{-(\frac{t}{CR} )} = 0.25

    Taking the logs of both sides

    -(\frac{t}{CR} ) = In(0.25)

    (\frac{t}{CR} ) = -In(0.25)

    t = =CR x In (0.25)

    t = (18 X 10^{-12} X 975) X (-1.386)

    t = 2.43 x 10∧-9

    t = 24.3ns

  2. kimchi2
    0
    2021-08-04T13:05:27+00:00
    Reply

    Answer:

    t = 23.9nS

    Explanation:

    given :

    Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

    distance d= 1mm=> 0.001

    resistor R= 975 ohm

    Capacitance can be calculated through the following formula,

    C = (ε0  x A )/d

    C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

    C = 17.7 x 10^-12    (pico ‘p’ = 10^-12)

    C = 17.7pF

    the voltage between two plates is related to time, There we use the following formula of the final voltage

    Vc = Vx (1-e^-(t/CR))  

    75 = 100 x (1-e^-(t/CR))

    75/100 = (1-e^-(t/CR))

    .75 = (1-e^-(t/CR))

    .75 -1 = -e^-(t/CR)

    -0.25 = -e^-(t/CR)  —>(cancelling out the negative sign)

    e^-(t/CR) = 0.25

    in order to remove the exponent, take logs on both sides  

    -t/CR = ln (0.25)

    t/CR = -ln(0.25)

    t = -CR x ln (0.25)

    t = -(17.7 x 10^-12 x 975) x (-1.38629)

    t = 23.9 x 10^{-9  

    t = 23.9ns

    Thus, it took 23.9ns  for the potential difference between the deflection plates to reach 75 volts

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