What is the speed of the moving charge when it's 1.0cm from the fixed charge?

[pat666]

Homework Statement

Two identical 25g particles each carry 5.0uC of charge. One is held fixed, and the other is placed 1.0mm away and released
Find the speed of the moving charge when it's 1.0cm from the fixed charge

Homework Equations

The Attempt at a Solution

I thought that using F=Q1Q2K/d^2 I could find the repelling force which is 2.25*10^5 N. then F=ma so a=2.25*10^5/0.025 =9*10^6m/s^2. then using kinematics v^2 =u^2+2as
0=u^2+2*9*10^6*0.009 the trouble is that that is unsolvable (imaginary solution) Its extremely possible that my procedure is entirely wrong, I was just making it up on the fly... Help!

 

[rock.freak667]

The Attempt at a Solution

I thought that using F=Q1Q2K/d^2 I could find the repelling force which is 2.25*10^5 N. then F=ma so a=2.25*10^5/0.025 =9*10^6m/s^2. then using kinematics v^2 =u^2+2as
0=u^2+2*9*10^6*0.009 the trouble is that that is unsolvable (imaginary solution) Its extremely possible that my procedure is entirely wrong, I was just making it up on the fly... Help!

The charge is held and then released, meaning that the initial velocity is zero.

 

[pat666]

argh thanks - stupid little mistake

 

[pat666]

Hey, I solved it but the answer I got (402.5m/s) is still wrong?

 

[rock.freak667]

Hey, I solved it but the answer I got (402.5m/s) is still wrong?

Recheck your force calculation, I don't think your force should be that high (in the order of 10⁵)

 

[pat666]

I rechecked it and got the same F=9E9*5E-6*5E-6/.001^2 = 2.25E5?

 

[rock.freak667]

I rechecked it and got the same F=9E9*5E-6*5E-6/.001^2 = 2.25E5?

So you happen to know the correct answer?

 

[pat666]

no its for one of those stupid online tests

 

[rock.freak667]

no its for one of those stupid online tests

I ask as my calculation gives around 300 m/s.

 

[pat666]

really that's significantly different to my answer - exact same procedure as me??

 

[pat666]

I just tried it again from scratch and i keep getting 402m/s can you tell me how you did it please

 

[rock.freak667]

really that's significantly different to my answer - exact same procedure as me??

Yes. Post what you did.

 

[pat666]

ok for the Force i got 2.25*10^5 N (F=KQ^2/.001^2)
then a = 2.25*10^5/.025 = 9E6m/s^2
v^2=0+2*9E6*0.009

 

[Dick]

v^2=u^2+2as only applies if the acceleration is constant. It's not. Use potential energy and energy conservation.

 

[pat666]

yeah I figured out to use E=K+U