Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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

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

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

Grade:12

1 Answers

Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
7 years ago
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

Think You Can Provide A Better Answer ?

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

ASK QUESTION

Get your questions answered by the expert for free