Step-by-Step Solution:

1. Identify Given Values and Uncertainties The given values for the capacitances and the voltage are: Capacitance 1: $C_1 = 2000 \pm 10$ pF Capacitance 2: $C_2 = 3000 \pm 15$ pF Voltage: $V = 5.00 \pm 0.02$ V

2. Calculate the Equivalent Capacitance ($C_{eq}$) For two capacitors connected in series, the equivalent capacitance $C_{eq}$ is given by: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$ This can be rearranged to: $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$ Substituting the nominal values: $C_{eq} = \frac{2000 \times 3000}{2000 + 3000} = \frac{6,000,000}{5000} = 1200 \text{ pF}$

3. Calculate the Error in the Equivalent Capacitance ($\Delta C_{eq}$) To find the error in $C_{eq}$, we use the formula for error propagation. Starting from the relation $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$, we differentiate it: $-\frac{dC_{eq}}{C_{eq}^2} = -\frac{dC_1}{C_1^2} - \frac{dC_2}{C_2^2}$ For maximum error, we add the absolute values of the individual errors, replacing differentials with deltas: $\frac{\Delta C_{eq}}{C_{eq}^2} = \frac{\Delta C_1}{C_1^2} + \frac{\Delta C_2}{C_2^2}$ Now, we solve for $\Delta C_{eq}$ and substitute the given values: $\Delta C_{eq} = C_{eq}^2 \left( \frac{\Delta C_1}{C_1^2} + \frac{\Delta C_2}{C_2^2} \right)$ $\Delta C_{eq} = (1200)^2 \left( \frac{10}{(2000)^2} + \frac{15}{(3000)^2} \right)$ $\Delta C_{eq} = (1.44 \times 10^6) \left( \frac{10}{4 \times 10^6} + \frac{15}{9 \times 10^6} \right)$ $\Delta C_{eq} = 1.44 \left( \frac{10}{4} + \frac{15}{9} \right) = 1.44 \left( 2.5 + \frac{5}{3} \right)$ $\Delta C_{eq} = 1.44 \left( \frac{7.5 + 5}{3} \right) = 1.44 \left( \frac{12.5}{3} \right) = 0.48 \times 12.5 = 6 \text{ pF}$

4. Calculate the Fractional Error in $C_{eq}$ The fractional error in the equivalent capacitance is: $\frac{\Delta C_{eq}}{C_{eq}} = \frac{6}{1200} = \frac{1}{200} = 0.005$

5. Determine the Formula for Percentage Error in Energy The energy $U$ stored in the capacitor combination is given by: $U = \frac{1}{2} C_{eq} V^2$ The fractional error in $U$, denoted as $\frac{\Delta U}{U}$, is found using the rules of error propagation for products and powers: $\frac{\Delta U}{U} = \frac{\Delta C_{eq}}{C_{eq}} + 2 \frac{\Delta V}{V}$ The constant $\frac{1}{2}$ has no error and does not contribute to the error calculation.

6. Calculate the Fractional Error in Voltage (V) $\frac{\Delta V}{V} = \frac{0.02}{5.00} = \frac{2}{500} = \frac{1}{250} = 0.004$

7. Calculate the Total Fractional Error in Energy (U) Substitute the fractional errors into the formula from Step 5: $\frac{\Delta U}{U} = 0.005 + 2(0.004)$ $\frac{\Delta U}{U} = 0.005 + 0.008 = 0.013$

8. Calculate the Percentage Error in Energy The percentage error is the fractional error multiplied by 100: $\text{Percentage Error} = \frac{\Delta U}{U} \times 100\%$ $\text{Percentage Error} = 0.013 \times 100\% = 1.3\%$

Thus, the percentage error in the calculation of the stored energy is 1.3.