How Close Does a Proton Get to an Infinitely Long Line of Charge?

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An infinitely long line of charge with a linear charge density of 6.50*10^-12 C/m creates an electric potential that affects a proton moving toward it. The proton, initially 17.5 cm away and traveling at 2900 m/s, will convert its kinetic energy into potential energy as it approaches the line. The relationship between the change in potential energy and kinetic energy can be expressed as q (V_{f} - V_{i}) = 1/2 m_{p} v^2. To find the potential at the starting point, integration can be used to derive a general expression. This approach will help determine how close the proton can get to the line of charge.
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Q An infinitely long line of charge has a linear charge density of 6.50*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 2900 m/s. How close does the proton get to the line charge? I am lost and do not know how to approach the problem or set it up.
 
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from taht distance what is the potential due to this infinite rod?
When the proton gets as close as it can all of its kinetic energy will be converted to potential energy
q (V_{f} - V_{i}) = \frac{1}{2} m_{p} v^2

Cna you find the potential at the starting point using integration? Try using variables so you cna get a general expresion. It will be useful later on.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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