Answer
Verified
430.5k+ views
Hint: The quantity of charge contained per unit area is called surface charge density. All charged conductors will experience some mechanical force. Elements having a surface charge density will always experience a mechanical force outwards. Here we have to derive an expression for the mechanical force.
Complete step by step answer:
Small elements of the conductor will experience a mechanical force normally outwards. Let us consider the small element $dS$ as shown below,
Let $\sigma $ be the surface charge density of the conductor. Then the charge carried by the small element $dS$ be $dq$.
Then we can write,
$dq = \sigma dS$ -------$(1)$
Let us consider a point $P$ outside the small element as shown in the figure.
At this point $P$, the electrical intensity can be written as,
$E = \dfrac{\sigma }{{{\varepsilon _0}k}}$ -------$(2)$
where ${\varepsilon _0}$ stands for the permittivity of free space and $k$ stands for the dielectric constant.
The electrical intensity is directed normally outwards.
The electrical intensity $E$ can be resolved into two components, ${E_1}$ and ${E_2}$.
$\overrightarrow {{E_1}} $ is the component of electrical intensity due to the small charge $dq$present in the element.
${\vec E_2}$ is the component due to the rest of the charges present on the surface.
Hence the total electrical intensity at the point $P$ can be written as,
$E = {E_1} + {E_2}$
Substituting this value in equation $(2)$
${E_1} + {E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$ --------$(3)$
Let us now consider the point $Q$. At this point, the components of electrical intensity are in opposite direction. Inside the charged conductor the total charge is zero.
i.e.
${E_1} - {E_2} = 0$
$\therefore {E_1} = {E_2}$
Substituting ${E_1} = {E_2}$ in equation $(3)$
We get
${E_2} + {E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$
This can be written as,
$2{E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$
From this, we can write ${E_2}$ as,
${E_2} = \dfrac{\sigma }{{2{\varepsilon _0}k}}$
We know that the electrical intensity due to the rest of the charges in the conductor is ${E_2}$. The element $dS$ of the conductor will experience a repulsive force due to this component.
The repulsive force experienced by the rest of the conductor due to the charge $dq$ in the small element $dS$ is given by,
$F = {E_2}dq$
We know that ${E_2} = \dfrac{\sigma }{{2{\varepsilon _0}k}}$
and $dq = \sigma dS$
Substituting these values in the equation for force,
$F = \dfrac{\sigma }{{2{\varepsilon _0}k}}.\sigma .ds = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}dS$
From this equation, we can write
$\dfrac{F}{{dS}} = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}$
The force per unit area can be written as, $\dfrac{F}{{dS}} = f$
Thus the above equation will become,
$f = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}$
This is the mechanical force per unit area of a conductor due to a surface charge density.
Note: The S.I unit for the mechanical force per unit area is given by $N/{m^2}$. Like charges repel each other and unlike charges attract. In a charge-carrying conductor, the charges are all similar. Hence when we consider a small element the rest of the conductor will experience the mechanical force due to the repulsion of like charges.
Complete step by step answer:
Small elements of the conductor will experience a mechanical force normally outwards. Let us consider the small element $dS$ as shown below,
Let $\sigma $ be the surface charge density of the conductor. Then the charge carried by the small element $dS$ be $dq$.
Then we can write,
$dq = \sigma dS$ -------$(1)$
Let us consider a point $P$ outside the small element as shown in the figure.
At this point $P$, the electrical intensity can be written as,
$E = \dfrac{\sigma }{{{\varepsilon _0}k}}$ -------$(2)$
where ${\varepsilon _0}$ stands for the permittivity of free space and $k$ stands for the dielectric constant.
The electrical intensity is directed normally outwards.
The electrical intensity $E$ can be resolved into two components, ${E_1}$ and ${E_2}$.
$\overrightarrow {{E_1}} $ is the component of electrical intensity due to the small charge $dq$present in the element.
${\vec E_2}$ is the component due to the rest of the charges present on the surface.
Hence the total electrical intensity at the point $P$ can be written as,
$E = {E_1} + {E_2}$
Substituting this value in equation $(2)$
${E_1} + {E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$ --------$(3)$
Let us now consider the point $Q$. At this point, the components of electrical intensity are in opposite direction. Inside the charged conductor the total charge is zero.
i.e.
${E_1} - {E_2} = 0$
$\therefore {E_1} = {E_2}$
Substituting ${E_1} = {E_2}$ in equation $(3)$
We get
${E_2} + {E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$
This can be written as,
$2{E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$
From this, we can write ${E_2}$ as,
${E_2} = \dfrac{\sigma }{{2{\varepsilon _0}k}}$
We know that the electrical intensity due to the rest of the charges in the conductor is ${E_2}$. The element $dS$ of the conductor will experience a repulsive force due to this component.
The repulsive force experienced by the rest of the conductor due to the charge $dq$ in the small element $dS$ is given by,
$F = {E_2}dq$
We know that ${E_2} = \dfrac{\sigma }{{2{\varepsilon _0}k}}$
and $dq = \sigma dS$
Substituting these values in the equation for force,
$F = \dfrac{\sigma }{{2{\varepsilon _0}k}}.\sigma .ds = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}dS$
From this equation, we can write
$\dfrac{F}{{dS}} = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}$
The force per unit area can be written as, $\dfrac{F}{{dS}} = f$
Thus the above equation will become,
$f = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}$
This is the mechanical force per unit area of a conductor due to a surface charge density.
Note: The S.I unit for the mechanical force per unit area is given by $N/{m^2}$. Like charges repel each other and unlike charges attract. In a charge-carrying conductor, the charges are all similar. Hence when we consider a small element the rest of the conductor will experience the mechanical force due to the repulsion of like charges.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Write the difference between order and molecularity class 11 maths CBSE
A rainbow has circular shape because A The earth is class 11 physics CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What are noble gases Why are they also called inert class 11 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between calcination and roasting class 11 chemistry CBSE