Problem Overview

We are solving for the charge $Q(t)$ in an RC circuit with the following parameters:

• Resistance: $R = 10$
• Capacitance: $C = 2$
• Voltage source: $E(t) = 0$
• Initial charge: $Q(0) = Q_0$

The governing equation for an RC circuit is:

$R \frac{dQ}{dt} + \frac{Q}{C} = E(t)$

Step 1: Simplify the Differential Equation

Given $E(t) = 0$, the equation becomes:

$R \frac{dQ}{dt} + \frac{Q}{C} = 0$

Substituting $R = 10$ and $C = 2$:

$10 \frac{dQ}{dt} + \frac{Q}{2} = 0$

Divide through by $10$:

$\frac{dQ}{dt} + \frac{Q}{20} = 0$

Step 2: Solve the Differential Equation

Rewriting:

$\frac{dQ}{dt} = -\frac{Q}{20}$

This is a separable differential equation. Rearrange and integrate:

$\int \frac{1}{Q} \, dQ = -\int \frac{1}{20} \, dt$

The solutions are:

$\ln|Q| = -\frac{t}{20} + C_1$

Exponentiate both sides:

$Q = e^{-\frac{t}{20} + C_1} = e^{C_1} e^{-\frac{t}{20}}$

Let $e^{C_1} = Q_0$ (initial charge):

$Q(t) = Q_0 e^{-\frac{t}{20}}$

Step 3: Determine Time to Reach 10% of Initial Value

We are solving for $t$ when $Q(t) = 0.1 Q_0$. Substituting into the equation:

$0.1 Q_0 = Q_0 e^{-\frac{t}{20}}$

Cancel $Q_0$ on both sides:

$0.1 = e^{-\frac{t}{20}}$

Take the natural logarithm:

$\ln(0.1) = -\frac{t}{20}$

Solve for $t$:

$t = -20 \ln(0.1)$

Step 4: Simplify $t$

Using $\ln(0.1) = \ln(1) - \ln(10) = -\ln(10)$, we have:

$t = -20 (-\ln(10)) = 20 \ln(10)$

Numerical approximation ($\ln(10) \approx 2.302$):

$t \approx 20 \times 2.302 = 46.04$

Final Results

1. The charge as a function of time is:

$Q(t) = Q_0 e^{-\frac{t}{20}}$

2. The time it takes for the charge to reach 10% of its initial value is approximately:

$t \approx 46.04 \, \text{seconds.}$

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Would you like further details or additional examples? Here are 5 related questions for exploration:

1. How does the time constant $\tau = RC$ relate to the decay rate of the charge?
2. What happens if $E(t)$ is a nonzero constant voltage instead of 0?
3. How would the solution change if the resistance $R$ were halved?
4. Can you derive the voltage $V_C(t)$ across the capacitor in this circuit?
5. What is the physical significance of $Q(t)$ decaying exponentially?

Tip:

In RC circuits, the time constant $\tau = RC$ is crucial for determining the rate of charge and voltage changes. The larger $\tau$, the slower the decay.