From the Gauss's Law, the electric field is very easy to find than by the Coulomb's law. Therefore the Gauss's theorem is used to determine the electric field, in this section.
A. Electric Field Intensity due to a Charged Sphere:
Let us consider a charged sphere having center (O), radius (R) and carrier charge (Q). In an isolated solid conducting sphere the charge is distributed uniformly over its outer surface.
i) When point lying outside (r > R):
From the figure, a point (P) is outside, at a distance (r) from the center of the sphere. Draw a sphere of radius (r) through the point (P) with the center (O), called Gaussian surface. This sphere has surface area: $A = 4 \pi r^2$.
If E is the electric field intensity at point P, the total electric flux passing through the Gaussian Surface is,
$\phi = E * A = E * 4 \pi r^2$ .......... (i)
According to the Gauss Theorem, we have
$\phi = \frac{Q}{\epsilon_0}$ ........... (ii)
From equation (i) and (ii)
∴ $E = \frac{Q}{4 \pi \epsilon_0 r^2}$ .......... (iii)
Thus, the field outside the sphere is the same, if whole of the charge were concentrated at the center of the sphere.
ii) When point lying on the surface of sphere (r = R):
The area of the Gaussian surface of radius (r = R) is equal to surface area of the charged sphere.
i.e. $A = 4 \pi R^2$
If $E$ is the electric field intensity at point (P), the total flux $\phi$ passing through the Gaussian surface is,
$\phi = E * A = E * 4 \pi r^2 = E * 4 \pi R^2$ .......... (i)
But from the Gauss's Law, the flux passing through the closed surface is
$\phi = \frac{Q}{\epsilon_0}$ .......... (ii)
From equation (i) and (ii)
∴ $E = \frac{Q}{4 \pi \epsilon_0 R^2}$ .......... (iii)
The electric field intensity at any point on the surface of the charged sphere is the same as through the whole charge were concentrated at the center of the sphere.
iii) When point lying inside the sphere (r < R):
The sphere of radius (r) is the Gaussian surface.
Since there is no charge inside the charged sphere, the Gaussian surface does not enclosed any charge. So the charge inside the Gaussian surface, $Q = 0$.
If $E$ is the electric field intensity at point (P), the total flux $\phi$ passing through the Gaussian surface is,
$\phi = E * A = E * 4 \pi r^2$ .......... (i)
But from the Gauss's Law, the flux passing through the closed surface is
$\phi = \frac{Q}{\epsilon_0}$ .......... (ii)
Where, charge (Q) inside the Gaussian surface is equal to zero.
From equation (i) and (ii), we have
$E = 0$ .......... (iii)
Thus, the electric field intensity is zero inside the charged sphere.
B. Electric Field Outside a Charged Plane Conductor:
Consider a plane conductor with the uniform surface charge density ($\sigma$), at a point (P) out side the conductor. The electric field intensity (E) at point (P) is perpendicular to the surface area (A) but parallel to the curved surface of the cylinder.
So, the flux passing normally through the surface area (A) is given by
$\phi = E * A$ .......... (i)
The net charged enclosed by the Gaussian Surface is $Q = \sigma * A$ .......... (ii)
From Gauss's theorem, the total flux passing through the cylinder is $\phi = \frac{Q}{\epsilon_0}$ .......... (iii)
From equation (i), (ii) and (iii)
$E = \frac{\sigma}{\epsilon_0}$ .......... (iv)
C. Electric Field Intensity due to Linear Charge Density:
Consider an infinitely long straight conductor of uniform linear charge density ($\lambda$ = charge per unit length). The charge enclosed by the Gaussian surface $Q = \lambda l$.
So, the flux passing normally through the surface area $(A = 2 \pi r l)$ is given by
$\phi = E * A = 2 \pi r l $ .......... (i)
The net charged enclosed by the Gaussian Surface is $Q = \lambda * l$ .......... (ii)
From Gauss's theorem, the total flux passing through the cylinder is $\phi = \frac{Q}{\epsilon_0}$ .......... (iii)
From equation (i), (ii) and (iii)
$E = \frac{\lambda}{2 \pi \epsilon_0 r}$ .......... (iv)
So, the electric field intensity depends on the linear charge density not on the total charge.
D. Electric Field of an Infinite Plane Sheet of Charge:
Consider a infinite plane sheet with a uniform surface charge density ($\sigma$). The flux crossing the side walls of the cylinder is zero beause E is parallel to the walls.
The electric flux passing through the two end flat faces is given by
$\phi = 2 * E A $ .......... (i)
The net charged enclosed by the Gaussian Surface is $Q = \sigma * A$ .......... (ii)
From Gauss's theorem, the total flux passing through the cylinder is $\phi = \frac{Q}{\epsilon_0}$ .......... (iii)
From equation (i), (ii) and (iii)
$E = \frac{\sigma}{2 \epsilon_0}$ .......... (iv)
Thus, for points near the sheet, the electric field intensity is independent of the distance from the sheet.
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