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: 12
        a particle of mass m and charge q is subjected to combined action  of gravity and a uniform horizontal electric field of strength E. it is projected from ground with speed v in the vertical plane parallel to the field at an angle  N to the horizontal . what is the maximum distance the object can travel horizontally before striking ground. the ans is v^2(qE +square root of (m^2g^2 +q^2E^2))  /mg^2
4 years ago

Answers : (1)

Sher Mohammad
IIT Delhi
askIITians Faculty
174 Points
							
force on particle
F= qe i+-mg k
intial velocity, u=VcosN i+ VsinN k
at maximum height vertical velocity is zero,
0=VsinN- mgt=0
t=V SinN/mg
the time to come back to ground = 2t= 2VSinN/mg
distance travelled , s=ut+.5 a t^2= VcosN*2VSinN/mg+ .5*qE*(2VSinN/mg)^2

sher mohammad
b.tech, iit delhi
4 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


Course Features

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


Course Features

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


Get Extra Rs. 1,590 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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