A particle of charge q and mass m is subjected to an electric field E = E₀(1 − ax²) in the x-direction, where a and E₀ are constants. Initially the particle was at rest at x = 0. Other then the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

Option 1 - <p>√(3/a)<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>√(1/a)</p>
Option 3 - <p>a</p>
Option 4 - <p>√(2/a)<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
3 Views|Posted 4 months ago
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1 Answer
V
4 months ago
Correct Option - 1
Detailed Solution:

E = E? (1 − ax²)

F = qE?
acceleration = F/m = (qE? /m) (1 - ax²) = v (dv/dx)
(qE? /m) ∫? (1 - ax²)dx = ∫? vdv; (qE? /M) (x - ax³/3) = 0
x (1 - ax²/3) = 0; x = 0 & x = √ (3/a)

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According to question, we can write

F C A = F C B = K q Q x 2 + ( d 2 ) 2 F = 2 F C A c o s θ = 2 K q Q x [ x 2 + ( d 2 ) 2 ] 3 2

For maxima of force

d F d x = 0 , s o

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