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 $15 p F$ ?
$(g i v e n ϵ_{0} = 8.86 \times 10^{- 12} C^{2} / N m^{2})$

Question

TestHub · Accepted Answer

Electric field inside capacitor = V d = V d A k ε 0 A k ε 0 = V C A k ε 0 ∵ k A ε 0 d = C 10 6 = 500 × 15 × 10 - 12 10 - 4 × k × 8.85 × 10 - 12 k = 15 × 500 8.85 × 10 - 4 × 10 6 = 75 8.85 k = 8.47 ≈ 8.5

Physics - Capacitor Question with Solution | TestHub

Options:

Doubts & Discussion