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This question involves charging a capacitor through a resistor, which is an RC (Resistor-Capacitor) charging circuit. Let’s walk through the steps to solve this.

### **Problem Setup:**
- Plate dimensions: 10 cm × 2 cm = \(10 \, \text{cm} \times 2 \, \text{cm} = 0.1 \, \text{m} \times 0.02 \, \text{m}\)
- Gap between plates: 1 mm = \(1 \times 10^{-3} \, \text{m}\)
- Potential difference applied: \(V_0 = 100 \, \text{V}\)
- Resistance: \(R = 975 \, \Omega\)
- Target voltage: \(V = 95 \, \text{V}\)
  
First, we need to calculate the capacitance \(C\) of the deflection plates, and then use the charging equation for the capacitor to determine the time required to reach 95 V.

---

### **Step 1: Calculate Capacitance (C)**
The formula for the capacitance of parallel plates is:

\[
C = \varepsilon_0 \frac{A}{d}
\]

Where:
- \(\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\) (permittivity of free space),
- \(A = 0.1 \times 0.02 = 0.002 \, \text{m}^2\) (area of one plate),
- \(d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\) (separation between plates).

\[
C = \frac{8.854 \times 10^{-12} \times 0.002}{1 \times 10^{-3}} = 1.77 \times 10^{-11} \, \text{F}
\]

---

### **Step 2: Use RC Circuit Charging Equation**
The voltage across a capacitor during charging follows:

\[
V(t) = V_0 \left(1 - e^{-t / RC}\right)
\]

Where:
- \(V(t)\) is the voltage at time \(t\),
- \(V_0\) is the initial applied voltage (100 V),
- \(R = 975 \, \Omega\),
- \(C = 1.77 \times 10^{-11} \, \text{F}\).

We need to solve for the time \(t\) when \(V(t) = 95 \, \text{V}\).

---

### **Step 3: Solve for Time (t)**
Rearrange the equation to isolate \(t\):

\[
\frac{V(t)}{V_0} = 1 - e^{-t / RC}
\]

\[
e^{-t / RC} = 1 - \frac{V(t)}{V_0}
\]

\[
e^{-t / RC} = 1 - \frac{95}{100} = 0.05
\]

Take the natural logarithm of both sides:

\[
-\frac{t}{RC} = \ln(0.05)
\]

\[
t = -RC \ln(0.05)
\]

---

### **Step 4: Substitute Values**
\[
t = -(975 \times 1.77 \times 10^{-11}) \ln(0.05)
\]

First, calculate \(RC\):

\[
RC = 975 \times 1.77 \times 10^{-11} = 1.727 \times 10^{-8} \, \text{seconds}
\]

Now, take the natural logarithm:

\[
\ln(0.05) \approx -2.996
\]

Finally:

\[
t = -(1.727 \times 10^{-8}) \times (-2.996)
\]

\[
t \approx 5.17 \times 10^{-8} \, \text{seconds}
\]

---

### **Answer:**
It takes approximately \(5.17 \times 10^{-8} \, \text{seconds}\) for the deflection plates to reach 95 volts.

---

Let me know if you have any questions or need more clarification!

---

**5 Related Questions:**
1. How would the time change if the resistance was doubled?
2. What happens to the charging time if the gap between the plates increases?
3. How does the area of the plates affect the capacitance?
4. Can you derive the charging equation from basic circuit laws?
5. How would the behavior change if the applied voltage was a sine wave instead of DC?

**Tip:** In RC circuits, the product \(RC\) is called the **time constant**, indicating the time for the voltage to reach about 63% of its final value.

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 975-ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95 volts?

Find out how to determine the time required for a capacitor in an RC circuit to reach 95 volts, given specific dimensions and resistance. This problem explores RC circuits, capacitance calculations, and exponential decay equations in physics.

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