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## Homework Statement

1.0 mA proton beam accelerated through potential difference of 1 keV.

Determine the volume charge density of the beam after acceleration assuming uniform current distribution within diameter of 5mm, with zero current outside of this.

Particle starting from rest.

Final answer I get is out by 10

^{-4}to the solution given in my notes. Pretty certain its a mistake in the notes, somebody seconding this would give me peace at mind.

## Homework Equations

ρ = I/νA

W = qV

ρ = charge density,

A = surface area of beam,

I = beam current,

v = proton velocity,

W = work

V = voltage

## The Attempt at a Solution

W = 1.6x10

^{-16}J = KE

_{final}(since starting from rest)

∴ 2W/m

_{proton}= v

^{2}

∴ v ≈4.4x10

^{5}

A = π 6.25x10

^{-6}m

^{2}

∴ ρ = 0.001 / π.(6.25x10

^{-6}).4.4x10

^{5}≈1.1x10

^{-4}