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 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?

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?

Grade:11

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

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 Experts
Badiuddin

 

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