A parallel-plate capacitor of plate area 40 cm^2 and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.

Question

Deepak Patra · Answer

Sol. A = 40 m2 = 40 * 10^–4 d = 0.1 mm = 1 * 10^–4 m R = 16 Ω ; emf = 2 V C = E base 0A/d = 8.85 * 10^-12 * 40 * 10^-4/1 * 10^-4 = 35.4 * 10^-11 F Now, E = Q/AE base 0 (1 – e^-t/RC) = 35.4 * 10^-11 */ 2/40 * 10^-4 * 8.85 * 10^-12 (1 – e^1.76) = 1.655 * 10^-4 = 1.7 * 10^-4 V/m.

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

A parallel-plate capacitor of plate area 40 cm^2 and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.

A parallel-plate capacitor of plate area 40 cm^2 and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Electromagnetic Induction

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship