To solve the problem step by step, we will use the formulas related to the energy stored in a capacitor and the relationship between energy, capacitance, and voltage.
### Step 1: Understand the given values
We are given:
- Maximum potential difference across the capacitor, \( V_{max} = 1.5 \, \text{V} \)
- Maximum energy stored in the capacitor, \( E_C = 360 \, \mu J = 360 \times 10^{-6} \, J \)
- Energy at a certain instant, \( E_{C'} = 40 \, \mu J = 40 \times 10^{-6} \, J \)
### Step 2: Use the energy formula for a capacitor
The energy stored in a capacitor is given by the formula:
\[
E_C = \frac{1}{2} C V^2
\]
Where:
- \( E_C \) is the energy stored in the capacitor,
- \( C \) is the capacitance,
- \( V \) is the potential difference across the capacitor.
### Step 3: Find the capacitance \( C \)
Using the maximum energy and maximum potential difference:
\[
360 \times 10^{-6} = \frac{1}{2} C (1.5)^2
\]
Calculating \( (1.5)^2 \):
\[
(1.5)^2 = 2.25
\]
Now substituting this back into the equation:
\[
360 \times 10^{-6} = \frac{1}{2} C (2.25)
\]
Rearranging to find \( C \):
\[
C = \frac{2 \times 360 \times 10^{-6}}{2.25}
\]
Calculating the right-hand side:
\[
C = \frac{720 \times 10^{-6}}{2.25} = 320 \times 10^{-6} \, F = 3.2 \, \mu F
\]
### Step 4: Use the energy formula for the instant energy
Now, we need to find the potential difference \( V' \) when the energy is \( 40 \, \mu J \):
\[
E_{C'} = \frac{1}{2} C (V')^2
\]
Substituting the known values:
\[
40 \times 10^{-6} = \frac{1}{2} (3.2 \times 10^{-6}) (V')^2
\]
Rearranging to solve for \( V' \):
\[
40 \times 10^{-6} = 1.6 \times 10^{-6} (V')^2
\]
Dividing both sides by \( 1.6 \times 10^{-6} \):
\[
\frac{40 \times 10^{-6}}{1.6 \times 10^{-6}} = (V')^2
\]
Calculating the left-hand side:
\[
\frac{40}{1.6} = 25
\]
Thus,
\[
(V')^2 = 25
\]
Taking the square root:
\[
V' = \sqrt{25} = 5 \, V
\]
### Final Answer
The potential difference across the capacitor at that instant is \( V' = 5 \, V \).
---
