A.) Electric field intensity at a different point in the field due to the uniformly charged solid conducting sphere:
Let us consider, A solid conducting sphere of radius $R$ in which $+q$ charge is distributed uniformly on the surface of the sphere. Now find the electric field intensity at different points due to the solid-charged conducting sphere. These different points are:
- Electric field intensity at an external point of the solid conducting sphere
- Electric field intensity on the surface of the solid conducting sphere
- Electric field intensity at an internal point of the solid conducting sphere
1. Electric field intensity at an external point of the solid conducting sphere:
If point O is the center of solid conducting spherical, The electric field at the outside of the sphere can be determined by the following steps →
- First, take the point $P$ outside the sphere
- Draw a spherical surface of radius r which passes through point $P$. This hypothetical surface is known as the Gaussian surface.
- Now take a small area $\overrightarrow {dA} $ around point $P$ on the Gaussian surface to find the electric flux passing through it.
- Now find the direction between the electric field vector and a small area vector.
Due to uniform charge distribution, the electric field intensity will be the same at every point on the Gaussian surface. So from the figure,
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| Electric field intensity due to a solid conducting sphere |
The direction of electric field intensity on the Gaussian surface is radially outward which is in the direction of the area vector of the Gaussian surface. i.e. ($\theta=0^{\circ}$). Here $\overrightarrow {dA}$ is a small area around point $P$ so the small electric flux $d\phi_{E}$ will pass through this small area $\overrightarrow {dA}$. so this flux can be found by applying Gauss's law in question given below:
$ d\phi_{E}= \overrightarrow {E}\cdot \overrightarrow{dA}$
$ d\phi_{E}= E\:dA\: cos\: 0^{\circ} \quad \left \{\because \theta=0^{\circ} \right \}$
$ d\phi_{E}= E\:dA \quad (1) \quad \left \{\because cos\:0^{\circ}=1 \right \}$
The electric flux passes through the entire Gaussian surface, So integrate the equation $(1)$ →
$\phi_{E}=\oint E\:dA\qquad (2)$
According to Gauss's law:
$ \phi_{E}=\frac{q}{\epsilon_{0}}\qquad (3)$
From equation (2) and equation (3), we can write as
$ \frac{q}{\epsilon_{0}}=\oint E\:dA$
$ \frac{q}{\epsilon_{0}}= E\oint dA$
The area of the entire Gaussian spherical is $\oint {dA}=4\pi r^{2}$. Now substitute this value in the above equation. So above equation can be written as:
$ \frac{q}{\epsilon_{0}}= E(4\pi r^{2})$
| $ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}$ |
From the above equation, we can conclude that the behavior of the electric field at the external point due to the uniformly charged solid conducting sphere is the same as, like the entire charge is placed at the center, point charge
If the surface charge density is $\sigma$, Then the total charge $q$ on the surface of a solid conducting sphere is→
$ q=4\pi R^{2}\: \sigma$
Substitute this value of charge $q$ in the above equation, so we can write the equation as:
$ E=\frac{1}{4\pi \epsilon_{0}}\frac{4\pi R^{2}\: \sigma}{r^{2}}$
| $ E=\frac{\sigma}{\epsilon_{0}}\frac{R^{2}}{r^{2}}$ |
This equation describes the electric field intensity at the external point of the solid conducting sphere.
2. Electric field intensity on the surface of the solid conducting sphere:
If point $P$ is placed on the surface of the solid conducting sphere i.e ($r=R$). so electric field intensity on the surface of the solid conducting sphere can be found by putting $r=R$ in the formula of electric field intensity at the external point of the solid conducting sphere:
| $ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{R^{2}}$ |
| $ E=\frac{\sigma}{\epsilon_{0}}$ |
3. Electric field intensity at an internal point of the solid conducting sphere:
If point $P$ is placed inside the solid conducting sphere then electric field intensity will be zero at that point because the charge is distributed uniformly on the surface of the solid sphere so there will not be any charge on the Gaussian surface and electric flux will be zero inside the solid sphere. i.e.
$ \phi_{E}=\oint E\:dA$
$ 0=E\oint dA \qquad\quad \left \{ \because \phi_{E}=0 \right \}$
Electric field intensity distribution with distance for Conducting Solid Sphere:
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| The electric field intensity distribution |
Electric field intensity distribution with distance shows that the electric field is maximum on the surface of the sphere and zero inside the sphere. Electric field intensity distribution outside the sphere reduces with the distance according to $E=\frac{1}{r^{2}}$.
B.) Electric field intensity at a different point in the field due to uniformly charged solid Non-conducting sphere:
Let us consider, A solid non-conducting sphere of radius R in which $+q$ charge is distributed uniformly in the entire volume of the sphere. So electric field intensity at a different point due to the solid charged non-conducting sphere:
- Electric field intensity at an external point of the solid non-conducting sphere
- Electric field intensity on the surface of the solid non-conducting sphere
- Electric field intensity at an internal point of the non-solid conducting sphere
1. Electric field intensity at an external point of the solid non-conducting sphere:
Let us consider, An external point $P$ which is at a distance $r$ from the center point $O$ of the sphere. The electric flux is radially outward in the sphere. So the direction of the electric field vector and the small area vector will be in the same direction i.e. ($\theta =0^{\circ}$). Here $\overrightarrow {dA}$ is a small area, the small amount of electric flux will pass through this area i.e. →
$ d\phi_{E}= \overrightarrow {E}\cdot \overrightarrow{dA}$
$ d\phi_{E}= E\:dA\: cos\: 0^{\circ} \quad \left \{\because \theta=0^{\circ} \right \}$
$ d\phi_{E}= E\:dA \qquad (1) \quad \left \{\because cos\:0^{\circ}=1 \right \}$
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| Electric field intensity due to the solid non-conducting sphere |
The electric flux passes through the entire Gaussian surface, So integrate the equation $(1)$ →
$ \phi_{E}=\oint E\:dA\qquad (2)$
According to Gauss's law:
$ \phi_{E}=\frac{q}{\epsilon_{0}}\qquad (3)$
From equation (1) and equation (2), we can write as
$ \frac{q}{\epsilon_{0}}=\oint E\:dA$
$ \frac{q}{\epsilon_{0}}= E\oint dA$
The area of the entire Gaussian spherical surface is $\oint {dA}=4\pi r^{2}$. Now substitute this value in the above equation. So the above equation can be written as:
$ \frac{q}{\epsilon_{0}}= E(4\pi r^{2})$
| $ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}$ |
From the above equation, we can conclude that the behavior of the electric field at the external point due to the uniformly charged solid non-conducting sphere is the same as the point charge i.e. like the entire charge is placed at the center.
Since the sphere is a non-conductor so the charge is distributed in the entire volume of the sphere. So charge distribution can calculate by volume charge density →
$q=\frac{4}{3} \pi R^{3} \rho $
Substitute this value of charge $q$ in the above equation, so we can write the equation as:
$ E=\frac{1}{4\pi \epsilon_{0}}\frac{4\pi R^{3}\: \rho}{3r^{2}}$
| $ E=\frac{\sigma}{\epsilon_{0}}\frac{R^{3}}{3r^{2}}$ |
This equation describes the electric field intensity at the external point of the solid non-conducting sphere.
2. Electric field intensity on the surface of a solid non-conducting sphere:
If point $P$ is placed on the surface of a solid non-conducting sphere i.e ($r=R$). so electric field intensity on the surface of a solid non-conducting sphere can be found by putting $r=R$ in the formula of electric field intensity at the external point of the solid non-conducting sphere:
| $ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{R^{2}}$ |
| $ E=\frac{\rho R}{3\epsilon_{0}}$ |
3. Electric field intensity at an internal point of the non-solid conducting sphere:
If point $P$ are placed inside the sphere and the distance from the origin $O$ is $r$, the electric flux which is passing through the Gaussian surface
$ \phi_{E}= E.4\pi r^{2}$
Where $\phi_{E}=\frac{q'}{\epsilon_{0}}$
$ \frac{q'}{\epsilon_{0}}=E.4\pi r^{2}$
Where $q'$ is part of charge $q$ which is enclosed with Gaussian Surface
$ E=\frac{1}{4 \pi \epsilon_{0}} \frac{q'}{r^{2}} \qquad \qquad (4)$
The charge is distributed uniformly in the entire volume of the sphere so volume charge density $\rho$ will be the same as the entire solid sphere i.e.
$ \rho=\frac{q}{\frac{4}{3}\pi R^{3}}=\frac{q'}{\frac{4}{3}\pi r^{3}}$
$ \frac{q}{\frac{4}{3}\pi R^{3}}=\frac{q'}{\frac{4}{3}\pi r^{3}}$
$ q'=q\frac{r^{3}}{R^{3}}$
$ q'=q\left (\frac{r}{R} \right)^{3}$
Put the value of $q'$ in equation $(4)$, so
$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}\left(\frac{r}{R} \right)^{3}$
$ E=\frac{1}{4\pi \epsilon_{0}}\frac{qr}{R^{3}}$
Where $q=\frac{4}{3} \pi R^{3} \rho $. So above equation can be written as:
$ E=\frac{1}{4\pi \epsilon_{0}}\frac{\frac{4}{3} \pi R^{3} \rho r}{R^{3}}$
| $E=\frac{ \rho r}{3 \epsilon_{0}}$ |
Electric field intensity distribution with distance for Non-conducting Solid Sphere:
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| Electric field intensity distribution due to solid non-conducting sphere |
Electric field intensity distribution with distance shows that the electric field is maximum on the surface of the sphere and zero at the center of the sphere. Electric field intensity distribution outside the sphere reduces with the distance according to $E=\frac{1}{r^{2}}$.
