An uncharged capacitor is connected in series with a resistor and a battery. The charging of the capacitor starts at t=0. The rate at which energy stored in the capacitor:-

Explanation

To solve the problem of determining the rate at which energy is stored in a capacitor connected in series with a resistor and a battery, we can follow these steps:

### Step 1: Understand the Circuit
We have a capacitor (C) connected in series with a resistor (R) and a battery (V). Initially, the capacitor is uncharged.

### Step 2: Write the Charge Equation
Using Kirchhoff's loop law, we can write the equation for the circuit:
\[ \frac{Q}{C} - IR = 0 \]
Where \( Q \) is the charge on the capacitor and \( I \) is the current through the circuit.

### Step 3: Relate Current to Charge
The current \( I \) can be expressed as the rate of change of charge:
\[ I = \frac{dQ}{dt} \]
Substituting this into the equation gives:
\[ \frac{Q}{C} - R\frac{dQ}{dt} = 0 \]
Rearranging this, we find:
\[ \frac{dQ}{dt} = \frac{Q}{RC} \]

### Step 4: Solve for Charge as a Function of Time
The solution to this differential equation is:
\[ Q(t) = CV(1 - e^{-t/(RC)}) \]
This describes how the charge on the capacitor increases over time.

### Step 5: Write the Energy Stored in the Capacitor
The energy \( U \) stored in the capacitor is given by:
\[ U = \frac{Q^2}{2C} \]
Substituting our expression for \( Q(t) \):
\[ U(t) = \frac{(CV(1 - e^{-t/(RC)})^2}{2C} \]
This simplifies to:
\[ U(t) = \frac{C V^2}{2}(1 - e^{-t/(RC)})^2 \]

### Step 6: Differentiate Energy with Respect to Time
To find the rate at which energy is stored, we differentiate \( U(t) \) with respect to time \( t \):
\[ \frac{dU}{dt} = \frac{d}{dt}\left(\frac{C V^2}{2}(1 - e^{-t/(RC)})^2\right) \]
Using the chain rule:
\[ \frac{dU}{dt} = CV^2(1 - e^{-t/(RC)})\left(\frac{d}{dt}(1 - e^{-t/(RC)})\right) \]
Calculating the derivative:
\[ \frac{d}{dt}(1 - e^{-t/(RC)}) = \frac{1}{RC} e^{-t/(RC)} \]
So,
\[ \frac{dU}{dt} = CV^2(1 - e^{-t/(RC)})\left(\frac{1}{RC} e^{-t/(RC)}\right) \]
This simplifies to:
\[ \frac{dU}{dt} = \frac{CV^2}{RC} (1 - e^{-t/(RC)}) e^{-t/(RC)} \]

### Step 7: Analyze the Behavior of the Rate of Energy Storage
As time \( t \) increases, \( e^{-t/(RC)} \) decreases from 1 to 0, meaning that \( (1 - e^{-t/(RC)}) \) increases from 0 to 1. Therefore, \( \frac{dU}{dt} \) will initially increase, reach a maximum, and then decrease as \( t \) approaches infinity.

### Conclusion
The rate at which energy is stored in the capacitor first increases and then decreases over time.