Calculating the Closest Distance of a Proton from an Infinite Line of Charge

AI Thread Summary
The discussion revolves around calculating the closest distance a proton can get to an infinite line of charge with a linear charge density of 8.00×10−12 C/m. The user initially attempts to use the electric field equation and force to determine acceleration and distance but realizes that the linear acceleration formula is only valid for constant acceleration. Suggestions are made to apply energy conservation principles instead, but confusion arises regarding the dependence of the electric field on distance from the line of charge. It is clarified that the electric field strength decreases with distance, and the user struggles to integrate the electric field or use potential energy effectively. Ultimately, the correct answer is stated to be 0.134 m, but the user remains uncertain about the calculations.
Ivegottheskill
Messages
11
Reaction score
0
The question I've been attempting:

An infinitely long line of charge has a linear charge density of 8.00×10−12 C/m. A proton is a distance of 17.5 cm from the line and moving directly toward the line with a speed of 2700 m/s.

How close does the proton get to the line of charge?
Use 1.60×10−19 C for the magnitude of the charge on an electron, 1.67×10−27 kg for the mass of a proton, and 8.85×10−12 F/m for the permittivity of free space

From my notes and working etc. I've got:

E = lambda/2*pi*epsilon*r (where lambda = charge per unit length and epsilon = permittivity of free space)

F = E * q (where q = the charge of the proton, -1.60*10^-19, as apparently defined by the question)

I've used the F calculated to get acceleration by

a = F/m (where m = mass of proton)

I get a value of -7.88*10^7

I then use the linear acceleration formula v^2 = u^2 + 2*a*s to try and calculate s.

I get 0.04627... (4.63*10^-2)

Apparently this is incorrect however. Can anyone see where I'm messing up?
 
Physics news on Phys.org
The linear acceleration formula holds only when the acceleration is constant.

Try using energy conservation instead! :)
 
Hmm, still having trouble,

1/2*m*v^2 + E*q*y = 0 + E*q*(unknown y value)

The formula for E that I'm using doesn't make sense to me but appears right in my notes and textbook: E = lambda (charge per unit length)/2*pi*r*epsilon(permittivity of free space)

I wouldn't think E relies on r (for a straight line of charge). Aren't field lines parallel for a line of charge? Meaning E is constant at any point in the field?

I'm not given the r value for the final state of the particle, so I can't work out either E or the unknown y on the right hand side of that formula.
 
The "Electric Field Lines" are spread out more (1-dim), farther from the wire.
That means the E-field strength decreases as 1/r .
You need substripts to distinguish "E_final" from "E_initial" ... not equal!

You want to integrate E(r) from y_initial to y_final ...
or if this isn't for calc-based physics, use Potential.
 
No, still confused out of my brain. Was doing it on Mastering Physics.com, but exceeded attempts and failed the question. I tried a billion random different formulas. Worst thing is having no idea if they were even valid to use in an equation

The answer was apparently 0.134 m, but I still can't see how. I'll probably have to see tutorial teacher or something :confused::confused:
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top