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Question

Physics Question on Electric charges and fields

A particle of charge qq and mass mm is subjected to an electric field E=E0(1ax2)E = E _{0}\left(1- ax ^{2}\right) in the xx -direction, where a and E0E_{0} are constants. Initially the particle was at rest at x=0x=0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

A

2a\sqrt{\frac{2}{a}}

B

1a\sqrt{\frac{1}{a}}

C

aa

D

3a\sqrt{\frac{3}{a}}

Answer

3a\sqrt{\frac{3}{a}}

Explanation

Solution

E=E0(1ax2)E = E _{0}\left(1- ax ^{2}\right)
W=qEdx=qE00x0(1ax2)dxW =\int qE\, dx = qE _{0} \int\limits_{0}^{ x _{0}}\left(1- ax ^{2}\right) dx
=qE0[x0ax033]= qE _{0}\left[ x _{0}-\frac{ ax _{0}^{3}}{3}\right]
For ΔKE=0,W=0\Delta KE =0, W =0
Hence x0=3ax _{0}=\sqrt{\frac{3}{ a }}