A disk of radius R with uniform positive charge density σ is placed on the x y plane with its center at the origin. The Coulomb potential along the z -axis is V z = σ 2 ϵ 0 R 2 + z 2 - z A particle of

Question

A disk of radius R with uniform positive charge density σ is placed on the x y plane with its center at the origin. The Coulomb potential along the z -axis is V z = σ 2 ϵ 0 R 2 + z 2 - z A particle of positive charge q is placed initially at rest at a point on the z -axis with z = z 0 and z 0 > 0 . In addition to the Coulomb force, the particle experiences a vertical force F → = - c k ^ with c > 0 . Let β = 2 c ε 0 q σ . Which of the following statement(s) is(are) correct?

TestHub · Accepted Answer

For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore,

$c z \geq \frac{σ q}{2 ϵ_{0}} [R - (\sqrt{R^{2} + z^{2}} - z)] \Rightarrow \frac{c 2 ϵ_{0}}{σ q} \geq \frac{[R - (\sqrt{R^{2} + z^{2}} - z)]}{z} \Rightarrow β \geq \frac{[R - (\sqrt{R^{2} + z^{2}} - z)]}{z}$

For option A

$\frac{[R - (\sqrt{R^{2} + z^{2}} - z)]}{z} = \frac{R - R (\sqrt{1 + \frac{625}{49}} - \frac{25}{7})}{R \frac{25}{7}} ≃ 0.242$

Since, $0.25 > 0.242$

Option A is correct.

For option B

$\frac{[R - (\sqrt{R^{2} + z^{2}} - z)]}{z} = \frac{R - R (\sqrt{1 + \frac{9}{49}} - \frac{3}{7})}{R \frac{3}{7}} ≃ 0.79$

Since, $0.25 < 0.79$

Option B is incorrect

For option C

$\frac{[R - (\sqrt{R^{2} + z^{2}} - z)]}{z} = \frac{R - R (\sqrt{1 + \frac{1}{3}} - \frac{1}{\sqrt{3}})}{R \frac{1}{\sqrt{3}}} ≃ 0.73$

Since, $0.25 < 0.73$

Therefore, the particle returns to $z_{0}$

Hence, option C is correct.

For option D,

For any value of $z > 0$

$\frac{[R - (\sqrt{R^{2} + z^{2}} - z)]}{z} < 1$

Hence, option D is correct.

Physics - Electrostatics Question with Solution | TestHub

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