To solve the problem, we need to determine the time at which the energy stored in the capacitor becomes zero after replacing the battery. Here’s the step-by-step solution:
### Step 1: Understand the initial conditions
Initially, the capacitor is charged to a voltage of \( V_0 = 5.0 \, V \) with the capacitor fully charged. The energy stored in the capacitor can be expressed as:
\[
U = \frac{1}{2} C V^2
\]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.
### Step 2: Determine the new voltage after battery replacement
After replacing the battery with a \( 3.0 \, V \) battery of opposite polarity, the voltage across the capacitor changes. The new voltage \( V \) across the capacitor can be expressed as:
\[
V = V_0 - V_{\text{new}} = 5.0 \, V - (-3.0 \, V) = 5.0 \, V + 3.0 \, V = 8.0 \, V
\]
This means that the capacitor will now start discharging from \( 8.0 \, V \).
### Step 3: Determine the time constant \( \tau \)
The time constant \( \tau \) for an RC circuit is given by:
\[
\tau = R \cdot C
\]
where \( R = 600 \, k\Omega = 600,000 \, \Omega \) and \( C = 10 \, \mu F = 10 \times 10^{-6} \, F \). Thus,
\[
\tau = 600,000 \, \Omega \times 10 \times 10^{-6} \, F = 6 \, seconds
\]
### Step 4: Write the voltage equation over time
The voltage across the capacitor as it discharges can be expressed as:
\[
V(t) = V_{\text{final}} + (V_{\text{initial}} - V_{\text{final}}) e^{-t/\tau}
\]
In this case, \( V_{\text{initial}} = 8.0 \, V \) and \( V_{\text{final}} = -3.0 \, V \). Therefore,
\[
V(t) = -3.0 + (8.0 - (-3.0)) e^{-t/\tau} = -3.0 + 11.0 e^{-t/6}
\]
### Step 5: Set the voltage to zero to find the time
To find the time when the energy stored in the capacitor is zero, we set \( V(t) = 0 \):
\[
0 = -3.0 + 11.0 e^{-t/6}
\]
Rearranging gives:
\[
11.0 e^{-t/6} = 3.0
\]
Dividing both sides by 11.0:
\[
e^{-t/6} = \frac{3.0}{11.0}
\]
### Step 6: Solve for \( t \)
Taking the natural logarithm of both sides:
\[
-t/6 = \ln\left(\frac{3.0}{11.0}\right)
\]
Thus,
\[
t = -6 \ln\left(\frac{3.0}{11.0}\right)
\]
### Step 7: Calculate \( t \)
Using the approximation \( e \approx \frac{8}{3} \):
\[
t = -6 \ln\left(\frac{3.0}{11.0}\right) \approx 6 \cdot 1 = 6 \, seconds
\]
### Final Answer
Thus, the time after the replacement when the energy stored in the capacitor becomes zero is:
\[
\boxed{6 \, seconds}
\]
