
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.
Answer
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Hint :To solve this question we have to know about the surface charge density. According to electromagnetism, surface charge density is known as the amount of electric charge per unit length, volume or surface area. Volume charge density is known as the charge per unit of the quantity that is measured in the SI system as coulombs per cubic meter at any given volume.
Complete Step By Step Answer:
According to the question,
Separation between the plates is $ 2.00cm $
Time taken by the electron is $ 2\mu s $
Mass of the electric is $ 9.1 \times {10^{ - 31}} $ jg
The charge of the electron is $ 1.6 \times {10^{ - 19}} $
We know that, permittivity of the vacuum is $ 8.84 \times {10^{ - 12}}f/m $
So, we can us here the formula,
$ f = eE $
The electric field is given by
$ E = \dfrac{\sigma }{{{\varepsilon _o}}} $
$ \therefore f = \dfrac{{e\sigma }}{{{\varepsilon _0}}} $
The distance covered by the accelerating charged particle is,
$ d = \dfrac{{a{t^2}}}{2} $
Now, using the equation $ f = ma $
$ \Rightarrow a = \dfrac{{2d}}{{{t^2}}} $
$ \dfrac{{e\sigma }}{{{\varepsilon _0}}} = \dfrac{{2dm}}{{{t^2}}} $
$ \sigma = \dfrac{{2dm{\varepsilon _0}}}{{e{t^2}}} $
This is the right answer.
Note :
We also have to keep in mind that, Surface charge describes the electric potential difference between the inner and outer surface of different states like solid and liquid, liquid and gas, or gas and liquid. The surface charge density is present only in conducting surfaces and describes the whole amount of charge q per unit area A. we know that the measurement for the accumulation of electric charge in a respective field is known as surface charge density. It helps in measuring the total quantity of electric charge as per the given dimension although dimension can be the area, length or volume of the electric body.
Complete Step By Step Answer:
According to the question,
Separation between the plates is $ 2.00cm $
Time taken by the electron is $ 2\mu s $
Mass of the electric is $ 9.1 \times {10^{ - 31}} $ jg
The charge of the electron is $ 1.6 \times {10^{ - 19}} $
We know that, permittivity of the vacuum is $ 8.84 \times {10^{ - 12}}f/m $
So, we can us here the formula,
$ f = eE $
The electric field is given by
$ E = \dfrac{\sigma }{{{\varepsilon _o}}} $
$ \therefore f = \dfrac{{e\sigma }}{{{\varepsilon _0}}} $
The distance covered by the accelerating charged particle is,
$ d = \dfrac{{a{t^2}}}{2} $
Now, using the equation $ f = ma $
$ \Rightarrow a = \dfrac{{2d}}{{{t^2}}} $
$ \dfrac{{e\sigma }}{{{\varepsilon _0}}} = \dfrac{{2dm}}{{{t^2}}} $
$ \sigma = \dfrac{{2dm{\varepsilon _0}}}{{e{t^2}}} $
This is the right answer.
Note :
We also have to keep in mind that, Surface charge describes the electric potential difference between the inner and outer surface of different states like solid and liquid, liquid and gas, or gas and liquid. The surface charge density is present only in conducting surfaces and describes the whole amount of charge q per unit area A. we know that the measurement for the accumulation of electric charge in a respective field is known as surface charge density. It helps in measuring the total quantity of electric charge as per the given dimension although dimension can be the area, length or volume of the electric body.
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