# 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

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1 year 2021-09-03T04:21:35+00:00 1 Answers 0 views 0

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

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