Recent content by Jon Blind

Sorry but you have to calculate the length from position 1 to position 2. It's an infinite linecharge. I can't understand where I went wrong in my calculations.

Yep I saw it, and I edited my post and did it ;) r2-r1=0.00237

In that case r2 should be= 0.00337m r2-r1=0.00237

Homework Statement q=1.602*10^-19 point 1 L=1mm=r1 v=1.1*10^6 at point 2 F=1.44*10^-12 at point 1 Homework Equations E=(1/4πε)*(q/r) ΔV=∫E*dr=(1/4πε)*q∫(1/r)=(1/4πε)*q*ln (r2/r1) ΔU=ΔK=mv^2/2 ΔK=mv^2/2=ΔV*q=q*(1/4πε)*Q*(ln(r2/r1))...

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

5.87*10^-4m THANKYOU! Freaking hell I was so confused

So Q=F/E ?? I'll give it a try and calculate it now, thankyou very much.

Yes, the mass of the proton is 1.673*10^-27 Epsilon=8.854*10^-12 and q=1.602*10^-19

Exactly and I'm trying to find out r2. According to my calculations r2=2.28*10^-13 but that seems way too little?

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...