Two capacitors with capacitance values C_1 = | vedclass

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(A) The energy stored in a series combination of capacitors is given by $U = \frac{1}{2} C_{eq} V^2$,where $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.First,

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$1.30$

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(A) The energy stored in a series combination of capacitors is given by $U = \frac{1}{2} C_{eq} V^2$,where $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.First,calculate the equivalent capacitance $C_{eq}$:$C_{eq} = \frac{2000 	imes 3000}{2000 + 3000} = \frac{6,000,000}{5000} = 1200 	ext{ pF}$.To find the error in $C_{eq}$,we use $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$. Differentiating,we get $\frac{\Delta C_{eq}}{C_{eq}^2} = \frac{\Delta C_1}{C_1^2} + \frac{\Delta C_2}{C_2^2}$.$\Delta C_{eq} = C_{eq}^2 \left( \frac{\Delta C_1}{C_1^2} + \frac{\Delta C_2}{C_2^2} ight) = (1200)^2 \left( \frac{10}{2000^2} + \frac{15}{3000^2} ight) = 1440000 \left( \frac{10}{4000000} + \frac{15}{9000000} ight) = 1440000 \left( 2.5 	imes 10^{-6} + 1.667 	imes 10^{-6} ight) \approx 6 	ext{ pF}$.Now,for energy $U = \frac{1}{2} C_{eq} V^2$,the relative error is $\frac{\Delta U}{U} = \frac{\Delta C_{eq}}{C_{eq}} + 2 \frac{\Delta V}{V}$.Percentage error $= \left( \frac{6}{1200} + 2 	imes \frac{0.02}{5.00} ight) 	imes 100 = (0.005 + 0.008) 	imes 100 = 1.3 \%$.

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