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Alexeev081 [22]
3 years ago
5

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

0 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
Physics
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer:

1.62\times 10^{-8}\ \text{s}

Explanation:

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

A = Area = 10\times 2\times 10^{-4}\ \text{m}^2

d = Distance between plates = 1 mm

V_c = Changed voltage = 60 V

V = Initial voltage = 100 V

R = Resistance = 1000\ \Omega

Capacitance is given by

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}

We have the relation

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}

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

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