Two capacitor with capacitance values `C_1 = 2000 om 10 pF and C_2 = 3000 pm 15pF` are connected in series. The voltage applied across this combination is `V = 5.00 pm 0.02V`. The percenntage arror in the calculation

Explanation

To solve the problem step by step, we need to find the percentage error in the energy stored in a series combination of two capacitors with given capacitance values and their respective errors.

### Step 1: Determine the equivalent capacitance of the capacitors in series.

The formula for the equivalent capacitance \( C \) of two capacitors \( C_1 \) and \( C_2 \) connected in series is given by:

\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]

Given:
- \( C_1 = 2000 \pm 10 \, \text{pF} \)
- \( C_2 = 3000 \pm 15 \, \text{pF} \)

Calculating the equivalent capacitance:

\[ \frac{1}{C} = \frac{1}{2000} + \frac{1}{3000} \]

Finding a common denominator (6000):

\[ \frac{1}{C} = \frac{3}{6000} + \frac{2}{6000} = \frac{5}{6000} \]

Thus,

\[ C = \frac{6000}{5} = 1200 \, \text{pF} \]

### Step 2: Calculate the error in the equivalent capacitance.

Using the formula for error propagation in series capacitance:

\[ \frac{dC}{C^2} = \left( \frac{dC_1}{C_1^2} + \frac{dC_2}{C_2^2} \right) \]

Where:
- \( dC_1 = 10 \, \text{pF} \)
- \( dC_2 = 15 \, \text{pF} \)

Calculating the errors:

\[ \frac{dC}{C^2} = \frac{10}{2000^2} + \frac{15}{3000^2} \]

Calculating each term:

\[ \frac{10}{4000000} + \frac{15}{9000000} = 0.0000025 + 0.0000016667 \approx 0.0000041667 \]

Now, multiply by \( C^2 \):

\[ dC = C^2 \cdot 0.0000041667 = (1200^2) \cdot 0.0000041667 \]

Calculating \( 1200^2 \):

\[ 1200^2 = 1440000 \]

Thus,

\[ dC \approx 1440000 \cdot 0.0000041667 \approx 6 \, \text{pF} \]

### Step 3: Calculate the energy stored in the capacitors.

The energy \( E \) stored in a capacitor is given by:

\[ E = \frac{1}{2} C V^2 \]

Given:

- \( V = 5.00 \pm 0.02 \, \text{V} \)

Calculating the energy:

\[ E = \frac{1}{2} \cdot 1200 \cdot (5)^2 = \frac{1}{2} \cdot 1200 \cdot 25 = 15000 \, \text{pJ} \]

### Step 4: Calculate the error in the energy.

Using error propagation for energy:

\[ \frac{dE}{E} = \frac{dC}{C} + 2 \frac{dV}{V} \]

Calculating \( \frac{dC}{C} \):

\[ \frac{dC}{C} = \frac{6}{1200} = 0.005 \]

Calculating \( \frac{dV}{V} \):

\[ \frac{dV}{V} = \frac{0.02}{5} = 0.004 \]

Thus,

\[ \frac{dE}{E} = 0.005 + 2 \cdot 0.004 = 0.005 + 0.008 = 0.013 \]

### Step 5: Calculate the percentage error in the energy.

To find the percentage error:

\[ \text{Percentage Error} = \frac{dE}{E} \cdot 100\% \]

Calculating:

\[ \text{Percentage Error} = 0.013 \cdot 100\% = 1.3\% \]

### Final Answer:
The percentage error in the calculation of the energy stored is **1.3%**.

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