Ratio of Kinetic Energies for Two Moving Charges in Uniform Electric Field

[Kishor Bhat]

Homework Statement

Two charges of masses m and 2m, charges 2q and q respectively are placed in a uniform electric field and are allowed to move at exactly the same time. Find the ratio of their kinetic energies.

Homework Equations

Field = qE
Force =(Kq1q2)/r^2
Kinetic energy=1/2mv^2

The Attempt at a Solution

I found their initial accelerations by finding the Coloumb force on each and then dividing by their respective masses. Then I used the equation v= u + at and found the kinetic energies. But I'm getting the ratio as 2:1 whereas the answer is 8:1.

 

[gneill]

It is entirely possible that the answer key is incorrect in this instance, and that you've found a correct result. Sometimes problems have values changed in order to make it a "new" problem, but updating the answer key is overlooked.

Why don't you try another approach to the problem and see if your result is the same? Hint: Perhaps use a conservation law?

 

[Kishor Bhat]

Hint: Perhaps use a conservation law?

I'm afraid I'm not getting it. Conservation of momentum gives 2:1, and for conserving energy, I don't know the distance between the charges to calculate potential energy.

 

[gneill]

I'm afraid I'm not getting it. Conservation of momentum gives 2:1, and for conserving energy, I don't know the distance between the charges to calculate potential energy.

Well, 2:1 confirms your earlier value, so it's strong evidence that the answer key is wrong and that your result is correct

 

[Kishor Bhat]

Ah. Well then, thank you. :)

 

[vela]

You're just misinterpreting the problem. You're assuming the force on each is due to the interaction of the two charges, but the problem apparently wants you to ignore that. Instead, the force on each is due to the uniform electric field imposed externally.

 

[gneill]

You're just misinterpreting the problem. You're assuming the force on each is due to the interaction of the two charges, but the problem apparently wants you to ignore that. Instead, the force on each is due to the uniform electric field imposed externally.

Huzzah! Well spotted Vela. I missed the uniform electric field bit in the problem statement and then assumed that it was just the two particles interacting. This changes the complexion of the problem significantly!

 

[Kishor Bhat]

But we don't know the magnitude of the electric field..

 

[gneill]

But we don't know the magnitude of the electric field..

You don't know the magnitude of q either, but that didn't stop you before

 

[Kishor Bhat]

So be it. If I use F = qE, and acceleration a = F/m, I get 8:1 since the velocities are proportional to a^2. But if we had to take the inter-particular forces into consideration, I couldn't use that formula, since the repulsive force would vary with intervening distance.

 

[gneill]

So be it. If I use F = qE, and acceleration a = F/m, I get 8:1 since the velocities are proportional to a^2. But if we had to take the inter-particular forces into consideration, I couldn't use that formula, since the repulsive force would vary with intervening distance.

Consider each of the particles separately interacting with the field. Remember, they have different masses and charges.

 

[vela]

So be it. If I use F = qE, and acceleration a = F/m, I get 8:1 since the velocities are proportional to a^2. But if we had to take the inter-particular forces into consideration, I couldn't use that formula, since the repulsive force would vary with intervening distance.

I think the question is just poorly written. Without more information, you can't really deal with the interaction between the two. For example, if they're really far apart, you might argue that the interaction could be ignored. If they're close to each other, their relative displacement relative to the direction of the electric field will affect the answer.