This question is about calculating the time it takes for the voltage across deflection plates in an oscilloscope to reach a certain value given an initial voltage and a resistor.

To solve this, let’s use the formula for charging a capacitor through a resistor in an RC (resistor-capacitor) circuit, where the voltage $V$ across the capacitor as a function of time $t$ is given by:

$V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)$

where:

• $V_0 = 100 \, \text{V}$ (initial applied voltage),
• $R = 1025 \, \Omega$ (resistor value),
• $C$ is the capacitance of the deflection plates.

Steps to solve:

1. Determine the capacitance $C$: Since the deflection plates are parallel, the capacitance $C$ can be calculated as: $C = \frac{\varepsilon_0 A}{d}$ where:

   • $\varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m}$ (permittivity of free space),
   • $A = 10 \, \text{cm} \times 2 \, \text{cm} = 0.1 \times 0.02 = 0.002 \, \text{m}^2$ (area of the plates),
   • $d = 1 \, \text{mm} = 0.001 \, \text{m}$ (distance between plates).

2. Calculate $C$ and then solve for $t$: Substitute the values into the capacitance formula, then use the RC circuit equation to find the time $t$ when $V(t) = 60 \, \text{V}$.

Would you like a step-by-step calculation to fully solve this?

Relative Questions:

1. How does the time constant $RC$ affect the charging rate of the capacitor?
2. What would happen if the resistance was increased or decreased?
3. Why is the exponential function used to model the charging behavior?
4. What role does the plate area play in determining capacitance?
5. How does a smaller gap between plates influence capacitance?

Tip: In RC circuits, the time constant $RC$ gives insight into how fast the capacitor charges to a certain percentage of the applied voltage.