NEET2019PhysicsCapacitanceActual
A capacitor of capacitance 9 nF having dielectric slab of _r=2.4 , dielectric strength 20 MV m ⁻¹ , and potential difference =20 ~V . Calculate area of plates.
Options
- A2.1 10⁻⁴ ~m ^2
- B4.2 10⁻⁴ ~m ^2
- C1.4 10⁻⁴ ~m ^2
- D2.4 10⁻⁴ ~m ^2
Correct answer
B. 4.2 10⁻⁴ ~m ^2
Step-by-step solution
Here, C=9 nF , _r=2.4, V=20 volt Dielectric strength =20 MV m ⁻¹ Let separation between plates =d aligned & E= V d 20 10^6= 20 d & d=10⁻⁶ ~m aligned Now, C= ₀ A _r d aligned & 9 10⁻⁹= 8.85 10⁻¹² A 2.4 10⁻⁶ & A= 9 10⁻¹⁵ 8.85 2.4 10⁻¹² =4.2 10⁻⁴ ~m ^2 aligned