Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
tanya tandon Grade: 11
`        A proton has an initial velocity of 3.82x10^7 m/s in the horizontal direction. It enters a uniform electric ?eld of 20000 N/C directed vertically. Ignoring gravitational e?ects, ?nd the time it takes the proton to travel 0.066 m horizontally. What is the vertical displacement of the pro- ton after the electric field acts on it for that time? What is the proton’s speed after being in the electric field for that time?`
7 years ago

## Answers : (1)

147 Points
```										Dear tanya
since no force in the horizontal direction so horiizonta velocity will reamain constant Vx = 3.82x107 m/s

so time taken to travel 0.066 m horixontally is  t = .066/3.82x107 = 1.73  X 10-9 s

charge of the proton is q = 1.6 * 10-19C
mass  of the proton is m = 1.67 * 10-27 kg
given E =20000 N/C
acceleration of proton in vertical direction  a=qE/m
=1.916 * 1012  m/sec2

vertical distance moved by proton  h= 1/2  at2
=1/2  * 1.916  * 1012 *(1.73  X 10-9)2
=2.867 * 10-6 m
vertical velocity of proton after time  t   Vy= at
=1.916  * 1012 *(1.73  X 10-9)
=3.315 * 103 m/sec
so resultanat velocity will be  = √Vx2 + Vy2

Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin

```
7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Mechanics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details