91Ó°ÊÓ

Fig. 23-25 shows an electron being released between two infinite non-conducting sheets and the table for three different situations is given.

Equating Newton's second law of motion with the force due to the electric field on a particle, the acceleration of the particle can be calculated. Now, considering the situation of the electric field for an infinite long conducting sheet, you can get the electric field value within the space of the sheets.

Formulae:

The electric force at a point due to a point charge,

$\mathrm{F}=\mathrm{qE}$ ….. (i)

The force due to Newton’s second law,

$\mathrm{F}=\mathrm{ma}$ ….. (ii)

The electric field for an infinite non-conducting sheet,

$\mathrm{E}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$ ….. (iii)

Here, $\mathrm{F}$ is the force, $\mathrm{q}$ is the charge, $E$ is the electric field, $m$ is the mass, $\mathrm{a}$ is the acceleration, $\mathrm{σ}$ is the density, and ${\mathrm{Î\mu }}_0$ is the permittivity of free space.

Using equation (iii), the electric field of the two non-conducting sheets can be given as:

${\mathrm{E}}_{+}=\frac{{\mathrm{σ}}_{+}}{2{\mathrm{Î\mu }}_0}$

${\mathrm{E}}_{−}=\frac{{\mathrm{σ}}_{−}}{2{\mathrm{Î\mu }}_0}$

Here, the direction of the electric fields is the same for both the sheets as the electron is attracted to the negative sheet and repulsed by the positive sheet (one of the electric fields has a negative sign).

Thus, the net electric field can be given as:

$\mathrm{E}=\frac{{\mathrm{σ}}_{+}}{2{\mathrm{Î\mu }}_0}−\frac{{\mathrm{σ}}_{−}}{2{\mathrm{Î\mu }}_0}$ ….. (iv)

Now, comparing equations (i) and (ii), we can get the acceleration of the electron as follows:

$\begin{array}{c}\mathrm{ma}=\mathrm{qE}\\ \mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}\end{array}$

$\mathrm{a}=\frac{\mathrm{q}}{\mathrm{m}}\left(\frac{{\mathrm{σ}}_{+}}{2{\mathrm{Î\mu }}_0}−\frac{{\mathrm{σ}}_{−}}{2{\mathrm{Î\mu }}_0}\right)$ ….. (v)

Now, for the given situation 1, the electron’s acceleration can be given using the given data from table in equation (v) as follows:

$\begin{array}{c}{\mathrm{a}}_1=\frac{\mathrm{q}}{\mathrm{m}}\left(\frac{4\mathrm{σ}}{2{\mathrm{Î\mu }}_0}−\left(\frac{−4\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\right)\right)\\ =\frac{\mathrm{q}}{\mathrm{m}}\left(\frac{8\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\right)\\ =\frac{4\mathrm{\pm çσ}}{{\mathrm{³¾Î\mu }}_0}\end{array}$

For the given situation 2, the electron’s acceleration can be given using the given data from table in equation (v) as follows:

$\begin{array}{c}{\mathrm{a}}_2=\frac{\mathrm{q}}{\mathrm{m}}\left(\frac{7\mathrm{σ}}{2{\mathrm{Î\mu }}_0}−\left(\frac{−\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\right)\right)\\ =\frac{\mathrm{q}}{\mathrm{m}}\left(\frac{8\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\right)\\ =\frac{4\mathrm{\pm çσ}}{{\mathrm{³¾Î\mu }}_0}\end{array}$

For the given situation 3, the electron’s acceleration can be given using the given data from table in equation (v) as follows:

$\begin{array}{c}{\mathrm{a}}_3=\frac{\mathrm{q}}{\mathrm{m}}\left(\frac{3\mathrm{σ}}{2{\mathrm{Î\mu }}_0}−\left(\frac{−6\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\right)\right)\\ =\frac{\mathrm{q}}{\mathrm{m}}\left(\frac{8\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\right)\\ =\frac{4\mathrm{\pm çσ}}{{\mathrm{³¾Î\mu }}_0}\end{array}$

Hence, the rank of the situations according to acceleration is ${\mathrm{a}}_1={\mathrm{a}}_2={\mathrm{a}}_3$.