Two capacitors with capacitance values $$C_1 = 2000 \pm 10$$ pF and $$C_2 = 3000 \pm 15$$ pF are connected in series. The voltage applied across this combination is $$𝑉 = 5.00 \pm 0.02$$ V. The percentage error in the calculation of the energy stored in this combination of capacitors is _______.

The energy stored in a capacitor (or an equivalent combination of capacitors) is given by $$E=\dfrac12\,C_{\text{eq}}\,V^2$$.

Hence, for percentage (fractional) errors, the rule for a product/quotient gives
$$\frac{\Delta E}{E}= \frac{\Delta C_{\text{eq}}}{C_{\text{eq}}}+2\,\frac{\Delta V}{V}\;.$$

We therefore need the percentage error in the equivalent capacitance $$C_{\text{eq}}$$ of two capacitors connected in series.

Step 1: Percentage errors in the individual capacitances
For $$C_1=2000\;\text{pF}\pm10\;\text{pF}$$
$$\frac{\Delta C_1}{C_1}= \frac{10}{2000}=0.005=0.5\%$$

For $$C_2=3000\;\text{pF}\pm15\;\text{pF}$$
$$\frac{\Delta C_2}{C_2}= \frac{15}{3000}=0.005=0.5\%$$

Step 2: Percentage error in $$C_{\text{eq}}$$ for a series combination
For two capacitors in series,
$$C_{\text{eq}}=\frac{C_1C_2}{C_1+C_2}\;.$$
Treating maximum (worst-case) errors, write it as a product/quotient:
$$C_{\text{eq}}=(C_1)(C_2)\bigl(C_1+C_2\bigr)^{-1}\;.$$
Thus, the fractional error is the sum of fractional errors of all three factors:
$$\frac{\Delta C_{\text{eq}}}{C_{\text{eq}}}= \frac{\Delta C_1}{C_1}+ \frac{\Delta C_2}{C_2}+ \frac{\Delta\!(C_1+C_2)}{(C_1+C_2)}\;.$$

The uncertainty in the sum is
$$\Delta\!(C_1+C_2)=\Delta C_1+\Delta C_2=10+15=25\;\text{pF}\;,$$
so
$$\frac{\Delta\!(C_1+C_2)}{C_1+C_2}= \frac{25}{2000+3000}= \frac{25}{5000}=0.005=0.5\%.$$

Therefore,
$$\frac{\Delta C_{\text{eq}}}{C_{\text{eq}}}=0.5\%+0.5\%+0.5\%=1.5\%.$$

Step 3: Percentage error in the applied voltage squared
Given $$V=5.00\;\text{V}\pm0.02\;\text{V},$$
$$\frac{\Delta V}{V}= \frac{0.02}{5.00}=0.004=0.4\%.$$
Because energy depends on $$V^2,$$ the contribution doubles:
$$2\,\frac{\Delta V}{V}=2\times0.4\%=0.8\%.$$

Step 4: Total percentage error in energy
$$\frac{\Delta E}{E}=1.5\%+0.8\%=2.3\%.$$

Hence, the percentage error in the calculated energy stored in the series combination of the two capacitors is 2.3 %.