JEE Main 20198 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2019
- Date
- 8 April
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Capacitance
JEE Main 2019 · 8 April · Morning Shift · Physics · Capacitance
Voltage rating of a parallel plate capacitor is 500V . Its...
Voltage rating of a parallel plate capacitor is Its dielectric can withstand a maximum electric field of The plate area is . What is the dielectric constant if the capacitance is ?
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
Electric field inside capacitor
\left(∵ \dfrac{kA{ε}_{0}}{d} = C\right){10}^{6} = \dfrac{500 \times 15 \times {10}^{- 12}}{{10}^{- 4} \times k \times 8.85 \times {10}^{- 12}}k = \dfrac{15 \times 500}{8.85 \times {10}^{- 4} \times {10}^{6}} = \dfrac{75}{8.85}k = 8.47 \approx 8.5
\left(∵ \dfrac{kA{ε}_{0}}{d} = C\right){10}^{6} = \dfrac{500 \times 15 \times {10}^{- 12}}{{10}^{- 4} \times k \times 8.85 \times {10}^{- 12}}k = \dfrac{15 \times 500}{8.85 \times {10}^{- 4} \times {10}^{6}} = \dfrac{75}{8.85}k = 8.47 \approx 8.5