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# Two large conducting plates are placed parallel to each other with a separation of $2.00cm$ between them. An electron starting from rest near one of the plates reaches the other plate in $2.00$ microseconds. Find the surface charge density on the inner surface.

Last updated date: 20th Jun 2024
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Hint: The motion of the electron is caused by the electric field between the plates and the force pulling the electron is electrostatic force between the two plates. Calculate the value of the acceleration and apply in the formula for the electrostatic force of attraction to find the value of the surface charge density.

It is given that an electron is travelling from one plate to the other which are separated by a distance of $2.00cm$.
Distance between the two plates, $u = 2.00cm = 2 \times {10^{ - 2}}m$
Time taken by the electron to travel from one plate to other, $t = 2\mu s = 2 \times {10^{ - 6}}s$
The field between the plates and the electrostatic force causing the electron to travel between the two plates.
Here we are asked to find the value of the surface charge density on the inner surface,
It is known that, the electric field, $E = \dfrac{\sigma }{{{\varepsilon _0}}}$
Where, $\sigma$ is the surface charge density
${\varepsilon _0}$ is the permittivity of the vacuum=$8.85 \times {10^{ - 12}}F{m^{ - 1}}$
We know electrostatic force, $F = qE$
Where,$q$ is the charge of electron = $1.6 \times {10^{ - 19}}c$
Applying the value of the electric field, we get,
$\Rightarrow F = q\dfrac{\sigma }{{{\varepsilon _0}}}$...........................equation (1)
But we know the force given by mass is multiplied by the acceleration.
That is, $F = {m_e} \times a$............................equation (2)
Where, ${m_e}$ is the mass of the electron = $9.1 \times {10^{ - 31}}kg$
$a$ is the acceleration
Acceleration is given by, $a = \dfrac{{2u}}{{{t^2}}}$
So equation 2 will become, $F = {m_e} \times \dfrac{{2u}}{{{t^2}}}$.................equation (3)
That is, from equations 1 and 3 we get,
$\Rightarrow F = q\dfrac{\sigma }{{{\varepsilon _0}}} = {m_e} \times \dfrac{{2u}}{{{t^2}}}$
$\Rightarrow \sigma = \dfrac{{{m_e}2u{\varepsilon _0}}}{{{t^2}q}}$
Applying all the known values we get,
$\Rightarrow \sigma = \dfrac{{9.1 \times {{10}^{ - 31}} \times 2 \times 2 \times {{10}^{ - 2}} \times 8.85 \times {{10}^{ - 12}}}}{{{{\left( {2 \times {{10}^{ - 6}}} \right)}^2} \times 1.6 \times {{10}^{ - 19}}}}$
$\therefore \sigma = 0.503 \times {10^{ - 12}}$
That is the surface charge density on the inner surface, $\sigma = 0.503 \times {10^{ - 12}}$.

Note: The electrostatic force or Coulomb force is capable of acting through space, producing an effect even when there is no physical contact between the particles involved. Since there is no physical contact required for this interaction, it is assumed that there exists an electric field around a charged particle. When another charged particle enters this electric field, forces of an electrical nature are formed.