Answer:
Incomplete question
This is the complete question
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 V potential difference is suddenly applied to the initially uncharged plates through a 1000 Ω resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95 V?
Explanation:
Given that,
The dimension of 10cm by 2cm
0.1m by 0.02m
Then, the area is Lenght × breadth
Area=0.1×0.02=0.002m²
The distance between the plate is d=1mm=0.001m
Then,
The capacitance of a capacitor is given as
C=εoA/d
Where
εo is constant and has a value of
εo= 8.854 × 10−12 C²/Nm²
C= 8.854E-12×0.002/0.001
C=17.7×10^-12
C=17.7 pF
Value of resistor resistance is 1000ohms
Voltage applied is V = 100V
This Is a series resistor and capacitor (RC ) circuit
In an RC circuit, voltage is given as
Charging system
V=Vo[1 - exp(-t/RC)]
At, t=0, V=100V
Therefore, Vo=100V
We want to know the time, the voltage will deflect 95V.
Then applying our parameters
V=Vo[1 - exp(-t/RC)]
95=100[1-exp(-t/1000×17.7×10^-12)]
95/100=1-exp(-t/17.7×10^-9)
0.95=1-exp(-t/17.7×10^-9)
0.95 - 1 = -exp(-t/17.7×10^-9)
-0.05=-exp(-t/17.7×10^-9)
Divide both side by -1
0.05=exp(-t/17.7×10^-9)
Take In of both sides
In(0.05)=-t/17.7×10^-9
-2.996=-t/17.7×10^-9
-2.996×17.7×10^-9=-t
-t=-53.02×10^-9
Divide both side by -1
t= 53.02×10^-9s
t=53.02 ns
The time to deflect 95V is 53.02nanoseconds
