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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

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Boffeemadrididn

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].

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