MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Radhika Batra Grade: 11
        A parallel-plate capacitor has plate area 20 cm^2 , plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.
3 years ago

Answers : (1)

Kevin Nash
askIITians Faculty
332 Points
										Sol. A 20 cm^2, d = 1 mm, K = 5, e = 6 V

R = 100 * 10^3 Ω, t = 8.9 * 10^-5 s
C = KE base 0A/d = 5 * 8.85 * 10^-12 *20 * 10^-4/1 * 10^-3
= 10 * 8.85 * 10^-3 * 10^-12/10^-3 = 88.5 * 10^-12
q = EC(1 – e^-tRC)
= 6 * 88.5 * 10^-12 (1 – e^-89*10^-6/88.5*10^-12*10^4) = 530.97
Energy = ½ * 500.97 * 530/88.5 * 10^-12
= 530.97 * 530.97/88.5 * 2 *10^12
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
  • Electricity and Magnetism
  • OFFERED PRICE: Rs. 1,696
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details