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 50 volts?

Answer 1

Answer:Resistor-Capacitor (RC) circuits, when driven by a voltage/current source, display a type of time-dependent charging or discharging since the charge from the capacitor goes through the resistor. When considering a discharging phase, the time-dependent voltage takes the form

Explanation:

Vt = Voe^-t/RC

Vt equal time dependent voltage

R is the resistance

C is the capacitance of the oscilloscope

Given

Area of capacitor equal to 0.1 * 0.002m

Distance between plates equal 1/1000 = 0.0001m

Vo = 100v supply voltage

Resistance R =:1000ohms

Vt time dependent voltage = 50V

To find capacitance C = Eo(A/d)

C = (8.85 * 10^-12)* (0.002/0.001)

C= 1.77 * 10^-11

Solving the equation Vt = Voe^-t/RC

t = 0.6931RC

t = 0.6931(1000)(1.77*10^-11)

t = 1.2*10^-9

t = 1.2ns

It will take the oscilloscope 1.2ns to reach 50V

Answer 2

It takes 12.27 ns for the voltage between the deflection plates to reach 50 volts.

We assume the setup is an RC circuit with the plates of the oscilloscope as the capacitor and the resistor in series with it.

First, we need to find the capacitance of the oscilloscope plate C from

C = ε₀A/d where ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m, A = area of plates = 10 cm by 2 cm = 20 cm² = 20 × 10⁻⁴ m² and d = gap distance = 1 mm = 1 × 10⁻³ m.

So, substituting the values of the variables into the equation, we have

C = ε₀A/d

C = 8.854 × 10⁻¹² F/m × 20 × 10⁻⁴ m²/1 × 10⁻³ m

C = 177.08 × 10⁻¹⁶ Fm/1 × 10⁻³ m

C = 177.08 × 10⁻¹³ F

C = 17.708  × 10⁻¹² F

Now, the voltage V across the deflection plates is given by

$V = V_{0}(1 - e^{-\frac{t}{RC} } )$ where V₀ = initial potential supplied = 100 V, R = resistance = 1000 Ω, t = time and C = capacitance of deflection plates = 17.708  × 10⁻¹² F

Since V = 50 V and we require the time it takes to reach 50 V, making t subject of the formula, we have

t = -RC㏑(1 - V/V₀)

Substituting the values of the variables into the equation, we have

t = -RC㏑(1 - V/V₀)

t = -1000 Ω × 17.708  × 10⁻¹² F㏑[1 - (50V/100V)]

t = -17.708  × 10⁻⁹ ㏑[1 - 0.5)]

t = -17.708  × 10⁻⁹ ㏑[0.5)]

t = -17.708  × 10⁻⁹ × -0.693

t = 12.274 × 10⁻⁹ s

t = 12.274 ns

t ≅ 12.27 ns

So, it takes 12.27 ns for the voltage between the deflection plates to reach 50 volts.

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