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

two charged particles if mass 4m and m having charges +q and +3q are placed in uniform electric field .they are aloowed to move for two seconds. find the ratio of their kinetic energies . assuming their own electric field intraction to each other is zero.

two charged particles if mass 4m and m having charges +q and +3q are placed in uniform electric field .they are aloowed to move for two seconds. find the ratio of their kinetic energies . assuming their own electric field intraction to each other is zero.

Grade:

3 Answers

bala s
4 Points
9 years ago

Since the question is ambiguous, I have assumed that the particles have been placed in the field at rest. We know the Coulombic force on each particle is the product of the charge on the particle and the electric field intensity, F=E*q.

Further, we can assume that the Electric field is due to a constant charge of magnitude Q.

Then, E=kQ/r2, where k is the Coulomb constant 9*109 N m2C-2

Thus, by calculating the Coulombic force for each particle, we can equate this to force acting on the particle according to Newtons Second Law.

F=ma, F=kqQ/r2 => ma=kqQ/r2. Subsequently we get an expression for acceleration of the body. Now, by assuming that it was initially at rest, we can use equations of 1 dimensional motion to get the expression for velocity after 2 seconds. Using this velocity in the expression for Kinetic Energy, K=mv2/2, we get expressions for Kinetic Energy of both particles. Dividing them and cancelling out the common terms, we get the ratio 1:36. Please refer to the image for detailed calculations.

1716_58024_2012-04-21 12.20.36.jpg

If my logic is incorrect or there are any errors in calculation or otherwise, please let me know.

bala s
4 Points
9 years ago

In the above reply, please right click the image and select "Open image in new tab" to view the full calculation and the answer.

Rishi Sharma
askIITians Faculty 646 Points
9 months ago
Dear Student,
Please find below the solution to your problem.

Let a particle of mass m have charge q and be accelerated from rest to speed v in a uniform electric field.
The electrostatic force exerted on the charged particle can be given as:
F = qE.
Therefore, acceleration of the particle is given by:
a = Fm = qEm.
The final velocity of the charged particle in time
t is given by:
v = at = qEmt.
Then kinetic energy of the particle in time t is given by:
K = 1/2*mv*2 = 1/2*m×(qEmt)^2 = q^2E^2t^2m/2.
Therefore, if the electric field and time are kept constant, then:
K ∝ q^2m
If we consider two charged particles of masses m1 and m2 and respective charges
q1 and q2, being accelerated in the same electric field and for the same interval of time, the ratio of their kinetic energies will be given by:
K1/K2 = (q1/q2)^2m1/m2

Given
The mass of the first particle:
m1 = 4m.
The charge of the first particle:
q1 = q.
The mass of the second particle
m2 = m
The charge of the second particle
q2 = 3q

Hence
K1/K2 = (q1/q2)^2 m1/m2
K1/K2 = (q/3q)^2 4m/m
K1/K2 = 4/9

Thanks and Regards

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