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
Grade: 12th pass
        
A THIN HALF RING OF RADIUS 20cm IS UNIFORMLY CHARGE WITH A TOTAL CHARGE Q=0.7nC.THE MAGNITUDE OF ELECTRIC FIELD AT THE CENTRE OF THE HALF RING IS
2 years ago

Answers : (1)

Vikas TU
9484 Points
							
We would need to calculate the electric field in both direction that is x and y respeectively.
Therefore for x direcn.,
dE = k’*k*dl*cos(thetha)/r^2
where k’ be the charge densite over the semi arc.
Now integerate both sides from theta 0 to pi.
O integerating we get,
Ex = (k*k’/r)(sinpi – sin0) = 0
Similarly in y direcn. we get,
dEy= k’*k*dl*cos(thetha)/r^2
Integerating both sides,
Ey = 2*K*K’/r
 
Net electric field => E = Ey => 2KK’/r
where K = 1/4pie
and K’ = 0.7 x 10^-9/pi*0.2.
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

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