Question

A large flat metal surface has a uniform charge density +σ+σ. An electron of mass m and charge (e) leaves the surface at point. A with speed u, and returns to it at
b. Neglecting gravity. The maximum value of ab is

Answer 1

Given:

The charge density is +б

To find:

The maximum value of ab is

Solution:

The formula of electric field on the charge density is

E=б/2ε₀

Now the range of ab is =U²/a

The value of  a is

a=ρE/m

Put the value of E back

a=ρб/2mε₀

Put the value of a in the range of ab formula

=2mε₀U²/ρб

That is the maximum value of ab

Answer 2

Answer:

AB = Range , $R = \frac{2u^2m\epsilon_0}{\sigma e}$

Explanation:

The given motion represents projectile motion and range is given by AB

Here Range,

$R = \frac{u^2 sin2\theta}{a}$

Where a is acceleration. Since, we do not account for gravity, the acceleration is provided by the electric force due to electric field by flat plate.

Field for a flat plate, E

$E = \frac{\sigma}{2\epsilon_0}$

Force = F

$F = qE\\\\ma = eE\\\\a = \frac{eE}{m}\\\\a =\frac{e \sigma}{2\epsilon_0 m}$

For maximum range, sin2θ = 1, as 2θ = 90⁰, θ = 45⁰

Thus range

$R = \frac{2u^2m\epsilon_0}{\sigma e}$