Two capacitors with capacitance values C _ 1 = (2000 ± 10)pF and C...

For the purpose of calculation of error, fundamental formula is considered

$\dfrac{1}{C} = \dfrac{1}{{C}_{1}} + \dfrac{1}{{C}_{2}} ⇒ C = 1200pF$

$- dC{C}_{1}^{2} = - d{C}_{1}{C}_{1}^{2} - d{C}_{2}{C}_{2}^{2}$

$dC = 6pF$

Equivalent capacitance $= \left(1200 \pm 6\right)pF$

$E = 1 / 2{CV}^{2}$

$( dE / E = dC / C + 2dV / V ) \times 100$

$= 1 . 3$