Question

he 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 do

Answer 1

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