Electromagnetism - The distance from point a to point b

[Jon Blind]

Homework Statement

So I want to know the distance to 2. The proton is at v=0 at the 1.

We know that:

q=1.602*10^-19 point 1

L=1mm

v=1.1*10^6 at point 2

F=1.44*10^-12 at point 1

Homework Equations

[/B]
E=(1/4πε)*(q/r²)

ΔV=∫E*dr=(1/4πε)*q∫(1/r²)=(1/4πε)*q*(1/r₂-1/r₁)

ΔU=ΔK=mv²/2

ΔK=mv²/2=ΔV*q=q*(1/4πε)*q*(1/r₂-1/r₁)

3) The attempt at a solutionI can't seem to calculate the distance. I don't know where I've gone wrong.

 

[Charles Link]

L=1 mm. Can you see that this is also $r_1$ ?

 

[Jon Blind]

Exactly and I'm trying to find out r2.

According to my calculations r2=2.28*10^-13 but that seems way too little?

 

[Charles Link]

What did you use for the mass of the proton? Also, did you convert $L$ to meters? Also you need to solve for Q. You can do that because itt tells you the force $F$ at point 1.

 

[Jon Blind]

Yes, the mass of the proton is 1.673*10^-27

Epsilon=8.854*10^-12

and q=1.602*10^-19

 

[Charles Link]

Yes, the mass of the proton is 1.673*10^-27

Epsilon=8.854*10^-12

and q=1.602*10^-19

I see one mistake. You assumed the two q's were equal. See also my edited previous post. You need to solve for $Q$.

 

[Jon Blind]

So Q=F/E ??

I'll give it a try and calculate it now, thankyou very much.

 

[Charles Link]

So Q=F/E ??

I'll give it a try and calculate it now, thankyou very much.

$F=\frac{Qq}{4 \pi \epsilon_o r^2}$. They give you $F$, $q$, and $r$. You need to compute $Q$.

 

[Jon Blind]

5.87*10^-4m

THANKYOU! Freaking hell I was so confused

 

[Charles Link]

5.87*10^-4m

THANKYOU! Freaking hell I was so confused

Compute $Q$ in Coulombs. You need this number for the remainder of the calculations. The answer you gave is incorrect.

 

[Jon Blind]

Compute $Q$ in Coulombs. You need this number for the remainder of the calculations. The answer you gave is incorrect.

How is that possible?

ΔK=mv2/2=ΔV*q=q*(1/4πε)*Q*(1/r2-1/r1)

Q=1.00*10^-9ΔK=mv2/2=ΔV*q=q*(1/4πε)*q*(1/r2-1/r1)

(mv^2*epsilon*m*4*pi)/(2*q*Q)=1/r2-1/r1

(1.673*10^-27)*)((1.1*10^6)^2)*4*pi*(8.854*10^-12)/(2*(1.602*10^-19)*(1.00*10^-9))=1/r2-1000

1/r2=1702.97

r2=5.872*10^-4

 

[Charles Link]

How is that possible?

ΔK=mv2/2=ΔV*q=q*(1/4πε)*Q*(1/r2-1/r1)

Q=1.00*10^-9ΔK=mv2/2=ΔV*q=q*(1/4πε)*q*(1/r2-1/r1)

(mv^2*epsilon*m*4*pi)/(2*q*Q)=1/r2-1/r1

(1.673*10^-27)*)((1.1*10^6)^2)*4*pi*(8.854*10^-12)/(2*(1.602*10^-19)*(1.00*10^-9))=1/r2-1000

1/r2=1702.97

r2=5.872*10^-4

Close, but your final term should read $\frac{1}{r_1}-\frac{1}{r_2} =1000-\frac{1}{r_2}$ . ( $r_2>r_1$). $\\$ Once you correctly solve for $r_2$, you then need to compute the distance $D=r_2-r_1$.

 

[Jon Blind]

In that case r2 should be= 0.00337m

r2-r1=0.00237

 

[Charles Link]

In that case r2 should be= 0.00337m

That's what I got also. :) :)

 

[Charles Link]

Now solve for $D$. See my edited post #12.

 

[Jon Blind]

Now solve for $D$. See my edited post #12.

Yep I saw it, and I edited my post and did it ;)

r2-r1=0.00237