A capacitor of capacitance C=2.0±0.1μF is charged to V=20±0.5V . The error in the

Q=CV=2.0×20=40 μC ΔQQ=ΔCC+ΔVV=0.12.0+(0.5)20=340⇒ΔQ=3μCU=12CV2=122202=400μJΔUU=ΔCC+2ΔVV=0.12.0+2(0.5)20=110⇒ΔU=40μJ

A capacitor of capacitance C=2.0±0.1μF is charged to V=20±0.5V . The error in the

Q=CV=2.0×20=40 μC ΔQQ=ΔCC+ΔVV=0.12.0+(0.5)20=340⇒ΔQ=3μCU=12CV2=122202=400μJΔUU=ΔCC+2ΔVV=0.12.0+2(0.5)20=110⇒ΔU=40μJ

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A capacitor of capacitance $C = (2.0 \pm 0.1) μ F$ is charged to $V = (20 \pm 0.5) V$ . The error in the

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charge stored in the capacitor is $6 μ C$

energy stored in the capacitor is $28 μ J$

charge stored in the capacitor is $5 μ C$

energy stored in the capacitor $40 μ J$

answer is D.

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Detailed Solution

$Q = CV = 2.0 \times 20 = 40 μC$

$\frac{ΔQ}{Q} = \frac{ΔC}{C} + \frac{ΔV}{V} = \frac{0.1}{2.0} + \frac{(0.5)}{20} = \frac{3}{40}$

$\Rightarrow Δ Q = 3 μC$

$U = \frac{1}{2} {CV}^{2} = \frac{1}{2} (2) {(20)}^{2} = 400 μJ$

$\frac{ΔU}{U} = \frac{ΔC}{C} + 2 \frac{ΔV}{V} = \frac{0.1}{2.0} + 2 \frac{(0.5)}{20} = \frac{1}{10}$

$\Rightarrow ΔU = 40 μJ$

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