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

Answer 1

`1.62* 10^(-8)\ \text{s}`

Step-by-step explanation:

`\epsilon_0` = Vacuum permittivity =
`8.854* 10^(-12)\ \text{F/m}`

`A` = Area =
`10* 2* 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=(\epsilon_0A)/(d)\\\Rightarrow C=(8.854* 10^(-12)* 10* 2* 10^(-4))/(1* 10^(-3))\\\Rightarrow C=1.7708* 10^(-11)\ \text{F}`

We have the relation

`V_c=V(1-e^{-(t)/(CR)})\\\Rightarrow e^{-(t)/(CR)}=1-(V_c)/(V)\\\Rightarrow -(t)/(CR)=\ln (1-(V_c)/(V))\\\Rightarrow t=-CR\ln (1-(V_c)/(V))\\\Rightarrow t=-1.7708* 10^(-11)* 1000\ln(1-(60)/(100))\\\Rightarrow t=1.62* 10^(-8)\ \text{s}`

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