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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorBasic capacitor analysis with dielectricEasy2 minQB

PhysicsEasysingle choice

Voltage rating of a parallel plate capacitor is $500 V .$ Its dielectric can withstand a maximum electric field of $10^{6} V / m .$ The plate area is $10^{- 4} m^{2}$ . What is the dielectric constant if the capacitance is $15 p F$ ?
$(g i v e n ϵ_{0} = 8.86 \times 10^{- 12} C^{2} / N m^{2})$

Options:

Answer:

Solution:

Electric field inside capacitor $= \frac{V}{d}$
$= \frac{V}{d} \frac{A k ε_{0}}{A k ε_{0}} = \frac{V C}{A k ε_{0}} (∵ \frac{k A ε_{0}}{d} = C)$
$10^{6} = \frac{500 \times 15 \times 10^{- 12}}{10^{- 4} \times k \times 8.85 \times 10^{- 12}}$
$k = \frac{15 \times 500}{8.85 \times 10^{- 4} \times 10^{6}} = \frac{75}{8.85}$
$k = 8.47 \approx 8.5$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Basic capacitor analysis with dielectric

⏱ 2mℹ️ Source: QB

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