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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 ______.

Solution
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$$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$$

$$C_{eq} = \dfrac{2000 \times 3000}{5000} = 1200 pF$$

$$\dfrac{-d C_{eq}}{C^2 _{eq}} = \dfrac{-d C_1}{C_1^2} - \dfrac{dC_2}{C_2^2}$$

$$dC_{aq} = 6PF$$

$$\varepsilon = \dfrac{1}{2} CV^2$$

$$\dfrac{d \varepsilon}{q} \times 100 = \left(\dfrac{dC}{C} + \dfrac{2dV}{V} \right) \times 100$$

$$= \left(\dfrac{6}{1200} + 2 \times \dfrac{0.02}{5}\right) \times 100$$

$$= 1.3\%$$

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