The plates of a parallel plate capacitor are charged to a potential difference of 117 V and then connected across a resistor. The potential difference across the capacitor decreases exponentially with to time.After 1s the potential difference between the plates is 39 V, then after 2s from the start, the potential difference (in V) between the plates is

Answer

To solve the problem, we need to analyze the behavior of the potential difference across a discharging capacitor over time. The potential difference decreases exponentially, and we can use the formula for the voltage across a discharging capacitor:

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

Where:
- \( V(t) \) is the potential difference at time \( t \)
- \( V_0 \) is the initial potential difference (117 V)
- \( R \) is the resistance
- \( C \) is the capacitance
- \( t \) is the time in seconds

### Step-by-step Solution:

1. **Identify the Initial Conditions**:
- Initial potential difference \( V_0 = 117 \, V \)
- At \( t = 1 \, s \), \( V(1) = 39 \, V \)

2. **Set Up the Exponential Decay Equation**:
- Using the formula for the voltage across the capacitor:
\[ V(1) = V_0 e^{-\frac{1}{RC}} \]
Substituting the known values:
\[ 39 = 117 e^{-\frac{1}{RC}} \]

3. **Solve for the Exponential Term**:
- Divide both sides by 117:
\[ \frac{39}{117} = e^{-\frac{1}{RC}} \]
Simplifying gives:
\[ \frac{1}{3} = e^{-\frac{1}{RC}} \]

4. **Take the Natural Logarithm**:
- Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{3}\right) = -\frac{1}{RC} \]

5. **Find the Voltage at \( t = 2 \, s \)**:
- Now we need to find \( V(2) \):
\[ V(2) = V_0 e^{-\frac{2}{RC}} \]
We can express this in terms of the previous result:
\[ V(2) = V_0 e^{-2 \cdot \left(-\ln\left(\frac{1}{3}\right)\right)} = V_0 \left(e^{-\frac{1}{RC}}\right)^2 \]
Substituting \( e^{-\frac{1}{RC}} = \frac{1}{3} \):
\[ V(2) = 117 \left(\frac{1}{3}\right)^2 = 117 \cdot \frac{1}{9} = 13 \, V \]

### Final Answer:
The potential difference between the plates after 2 seconds is **13 V**.