Answer
Verified
435k+ views
Hint A uniformly charged sphere is spherically symmetric i.e. all points around the sphere are identical. Therefore, choose a spherical Gaussian surface whose radius is equal to the distance$r$ from the center of the sphere.
Formula used: $\oint {\vec E \cdot d\vec A} = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$ where $\vec E$ is the electric field vector, $d\vec A$ is the infinitesimal area vector, ${Q_{enc}}$ is the charge enclosed in the Gaussian surface and ${\varepsilon _0}$ is the permittivity of free space, $\oint {\vec E \cdot d\vec A} $ is the total flux going through the Gaussian surface.
Complete step by step answer
Gauss law allows us to easily find the electric field of a charge distribution by taking advantage of a possible symmetry in its arrangement. From integral form of Gauss law, we have
$\Rightarrow \oint {\vec E \cdot d\vec A} = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$
For a uniform spherical charge, the equation becomes
$\Rightarrow E \times A = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$ because the electric fields flowing through every infinitesimal area of our Gaussian surface are equal.
Then,
$\Rightarrow E \times 4\pi {r^2} = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$ (since $A = 4\pi {r^2}$)
Rearranging for $E$, we get
$\Rightarrow E = \dfrac{{{Q_{enc}}}}{{4\pi {\varepsilon _0}{r^2}}}$
$\therefore E \propto \dfrac{1}{{{r^2}}}$
The electric field for a uniformly charged thin spherical shell at radius $r < R$ is zero (since there would be no charge enclosed).
Hence, the graph of electric field with distance is as shown below.
Note
For $r < R$ we said that this is due to the fact that there would be no charged enclosed. For further understanding consider the constant form of Gauss’s law
$\Rightarrow E \times A = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$
For $r < R$, we pick a Gaussian surface that is inside the spherical shell. Since the charges are distributed around the shell, the charge enclosed by the Gaussian surface is zero, i.e. $\Rightarrow {Q_{enc}} = 0$. Hence, from
$\Rightarrow E \times A = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}} $
$\Rightarrow E = 0 $
Also, $E$ is maximum at $r = R$ because ${Q_{enc}} = \sigma 4\pi {R^2}$ where $\sigma $ is the surface charge density. Thus, from $E = \dfrac{{{Q_{enc}}}}{{4\pi {\varepsilon _0}{r^2}}}$ we replace ${Q_{enc}}$ as $\sigma 4\pi {R^2}$. Then, we have
$\Rightarrow E = \dfrac{{\sigma 4\pi {R^2}}}{{4\pi {\varepsilon _0}{r^2}}}$.
For $r = R$,
$\Rightarrow E = \dfrac{\sigma }{{{\varepsilon _0}}}$
Formula used: $\oint {\vec E \cdot d\vec A} = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$ where $\vec E$ is the electric field vector, $d\vec A$ is the infinitesimal area vector, ${Q_{enc}}$ is the charge enclosed in the Gaussian surface and ${\varepsilon _0}$ is the permittivity of free space, $\oint {\vec E \cdot d\vec A} $ is the total flux going through the Gaussian surface.
Complete step by step answer
Gauss law allows us to easily find the electric field of a charge distribution by taking advantage of a possible symmetry in its arrangement. From integral form of Gauss law, we have
$\Rightarrow \oint {\vec E \cdot d\vec A} = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$
For a uniform spherical charge, the equation becomes
$\Rightarrow E \times A = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$ because the electric fields flowing through every infinitesimal area of our Gaussian surface are equal.
Then,
$\Rightarrow E \times 4\pi {r^2} = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$ (since $A = 4\pi {r^2}$)
Rearranging for $E$, we get
$\Rightarrow E = \dfrac{{{Q_{enc}}}}{{4\pi {\varepsilon _0}{r^2}}}$
$\therefore E \propto \dfrac{1}{{{r^2}}}$
The electric field for a uniformly charged thin spherical shell at radius $r < R$ is zero (since there would be no charge enclosed).
Hence, the graph of electric field with distance is as shown below.
Note
For $r < R$ we said that this is due to the fact that there would be no charged enclosed. For further understanding consider the constant form of Gauss’s law
$\Rightarrow E \times A = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$
For $r < R$, we pick a Gaussian surface that is inside the spherical shell. Since the charges are distributed around the shell, the charge enclosed by the Gaussian surface is zero, i.e. $\Rightarrow {Q_{enc}} = 0$. Hence, from
$\Rightarrow E \times A = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}} $
$\Rightarrow E = 0 $
Also, $E$ is maximum at $r = R$ because ${Q_{enc}} = \sigma 4\pi {R^2}$ where $\sigma $ is the surface charge density. Thus, from $E = \dfrac{{{Q_{enc}}}}{{4\pi {\varepsilon _0}{r^2}}}$ we replace ${Q_{enc}}$ as $\sigma 4\pi {R^2}$. Then, we have
$\Rightarrow E = \dfrac{{\sigma 4\pi {R^2}}}{{4\pi {\varepsilon _0}{r^2}}}$.
For $r = R$,
$\Rightarrow E = \dfrac{\sigma }{{{\varepsilon _0}}}$
Recently Updated Pages
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
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Who was the Governor general of India at the time of class 11 social science CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference Between Plant Cell and Animal Cell