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Question

A particle with specific charge qm moves rectilinearly due to an electric field E=E0ax, where a is a positive constant and x is the distance from the point where the particle was initially at rest. The distance covered by the particle till the moment it comes to standstill is nE0a.The value of n is

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Solution

Under the influence of electric field
Fe=qE=q(E0ax)
mdvdt=q(E0ax)
vdvdx=qm(E0ax)
vdv=qm(E0ax)dx
v22=qm(E0xax22)+C

Using initial condition,
At x=0, v=0 and C=0
v2=2qm[E0xax22]

When the particle comes to rest (i.e V = 0)
0=E0xax22

x=2E0a

x=2E0a is the distance covered by the particle till the moment it comes to rest.

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