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Question: A large flat metal surface has a uniform charge density \(\left( { + \sigma } \right)\). An electron...

A large flat metal surface has a uniform charge density (+σ)\left( { + \sigma } \right). An electron of mass m and charge ρ\rho leaves the surface at point A with speed u and returns to it at point B. Disregarding gravity, the maximum value of AB is:

A) 2μ2mε0σρ\dfrac{{2{\mu ^2}m{\varepsilon _0}}}{{\sigma \rho }}
B) μ2.ρε0μρ\dfrac{{{\mu ^2}.\rho {\varepsilon _0}}}{{\mu \rho }}
C) μ2ρε0mσ\dfrac{{{\mu ^2}\rho }}{{{\varepsilon _0}m\sigma }}
D) μ2σρε.m\dfrac{{{\mu ^2}\sigma \rho }}{{\varepsilon .m}}

Explanation

Solution

The electron releasing is a projectile and its trajectory (path) is parabolic due to the electric force acting downward on the electron because of the positively charged plate. The maximum value of AB can be decided by how far the electron can go when released from the surface which is given by its range.
R=u2sin⁡2θgR = \dfrac{{{u^2}\sin 2\theta }}{g}
Electric field (E) for uniform surface is σ2ε0\dfrac{\sigma }{{2{\varepsilon _0}}}
Other relationships that can be used are:
F = ma
F = qE

Complete step by step answer:
When the electron is released from point A, it will move in a parabolic path as the force on the electron will be downwards at all the points since the electron is negatively charged and the surface is positively charged. Diagrammatically:

The maximum value of AB is given by the maximum range of this projectile
Range (R) of a projectile is given as:
R=u2sin⁡2θgR = \dfrac{{{u^2}\sin 2\theta }}{g}
Maximum value of sin⁡θ\sin \theta = 1 at θ=90∘\theta = {90^ \circ }
But here we have sin⁡2θ\sin 2\theta , the angle will become:
2θ=90∘2\theta = {90^ \circ }
θ=45∘\theta = {45^ \circ }
Now, sin⁡2θ\sin 2\theta = 1
Maximum range is given as:
R=u2gR = \dfrac{{{u^2}}}{g}
But we have to disregard gravity (given in the question), so instead of considering acceleration due to gravity ‘g’ we will consider the acceleration due to electric force ‘a’. So,
R=u2aR = \dfrac{{{u^2}}}{a} _______ (1)
From newton’s second law of motion:
F = ma
a=Fma = \dfrac{F}{m} ____________ (2)
Force in terms of charge and electric field is:
F = qE ______________ (3)
Electric field for uniform surface is σ2ε0\dfrac{\sigma }{{2{\varepsilon _0}}}, charge of electron be e, equation (3) becomes:
F = eσ2ε0\dfrac{{e\sigma }}{{2{\varepsilon _0}}}
Substituting this value in (2), we get:
a=eσ2ε0ma = \dfrac{{e\sigma }}{{2{\varepsilon _0}m}}
Therefore, the acceleration due to electric force is eσ2ε0m\dfrac{{e\sigma }}{{2{\varepsilon _0}m}}
Replacing this value of a in (1) to get maximum range:
R=u2(eσ2ε0m)R = \dfrac{{{u^2}}}{{\left( {\dfrac{{e\sigma }}{{2{\varepsilon _0}m}}} \right)}}
R=2μ2mε0σρR = \dfrac{{2{\mu ^2}m{\varepsilon _0}}}{{\sigma \rho }}
Therefore, the maximum value of AB is 2μ2mε0σρ\dfrac{{2{\mu ^2}m{\varepsilon _0}}}{{\sigma \rho }} and thus the correct option is A).

Note: When an object is projected in the air, it is called a projectile and the path it covers is called trajectory.
The parabolic path is covered by a projectile when a downward force is acting on the body at all its points during the motion.
The range can be defined as the distance that the projectile covers