Question

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 65 volts?

Answer 1

Answer:

18.5 ns.

Explanation:

In order to develop the problem we must first identify the capacitance that is found

on the oscilloscope, like this:

$c =\frac{A \epsilon_ {0}}{d}$

Where A scope of the oscilloscope,

$\epsilon_ {0}$  is Vacuum permittivity

d = distance between the ones.

$C= (0.10m) (0.02m) (8.85 * 10 ^{- 12} C^ 2 / Nm ^ 2)/(1*10^{-3}m)$

$C = 17.7 * 10 ^{-12}F$

Defining the following variables in question we have to

Resistance (R) = 1000 Ohm

Meanwhile the Maximum Voltage (V_ {max}) applied is 100V

However, the maximum time to reach the voltage of 65V is

$V = V_ {max} (1- e ^{- t / (RC)})$

$65V = 100V (1-exp (-\frac{t}{1000 * 17.7 * 10^{-12}} )$

$0.65 = (1-exp (- \frac{t}{1000 * 17.7 * 10 ^{ -12}})$

$ln (0.35) =-\frac{t}{ 1000*17.7*10^{-12}}$

$t = - 1000 * 17.7 * 10 ^ {- 12} ln (0.35)$

t = 18.5ns