Finding the New Acceleration Given Point Charges

[Alex Z]

Homework Statement

A point charge q1=15.00μC is held fixed in space. From a horizontal distance of 8.00 cm , a small sphere with mass 4.00×10−3kg and charge q2=+2.00μC is fired toward the fixed charge with an initial speed of 40.0 m/s . Gravity can be neglected.
What is the acceleration of the sphere at the instant when its speed is 23.0 m/s?

2. The attempt at a solution
I tried to use F = (k*q*q) / (r^2) to find the force of the system, which comes out to be 41.72 N. But when I try to use a = F/m, the answer is apparently wrong.

 

[Doc Al]

How did you calculate the force? (Where is the sphere when its speed is 23.0 m/s?)

 

[Alex Z]

How did you calculate the force? (Where is the sphere when its speed is 23.0 m/s?)

I calculated the force using everything was given to me. Since the radius of the sphere wasn't given, I just assumed that the size of the sphere was negligible and I therefore didn't have to take it into account.
As for the location of the sphere, I assume conservation of energy needs to be used to find the new distance?

 

[Doc Al]

I calculated the force using everything was given to me. Since the radius of the sphere wasn't given, I just assumed that the size of the sphere was negligible and I therefore didn't have to take it into account.

Sounds good.

As for the location of the sphere, I assume conservation of energy needs to be used?

Sure. If you show your calculations, we can look for possible errors. (Or confirm your results.)

 

[Alex Z]

Well I believe that the equation here would be Ue_i + Ki = Ue_f + Kf
So (k*q*q)/0.08 + 3.44 = (k*q*q)/r_f + 1.137
With all this equated, r_f should be equal to 0.0487 meters.

From there, my first instinct would be to find the change in distance and then use kinematics.

 

[Doc Al]

Show how you calculated the kinetic energies. (Check your arithmetic.)

From there, my first instinct would be to find the change in distance and then use kinematics.

Once you have the final position, you won't need kinematics.

 

[Alex Z]

Initial Kinetic Energy = 1/2*(4*10^-3)*(40^2) = 3.2
Final Kinetic Energy = 1/2*(4*10^-3)*(23^2) = 1.058

Oops. So that means that r_f = 0.0487 meters. I guess the values weren't great enough to make a significant difference.

 

[Doc Al]

Now show how you calculated the force.

 

[Alex Z]

Now show how you calculated the force.

F = (k*q*q)/(r^2) = ((8.9*10^9)*(15*10^-6)*(2*10^-6)) / (0.0487^2) = 112.58 N

So from there, I can use F = m*a to find the acceleration:

a = F / m = (112.58) / (4*10^-3) = 28144.5 m/s^2

I believe that should be the final answer.

 

[Doc Al]

Much better!

 

[Alex Z]

Much better!

Awesome, thank you very much.