Derivation of electric field intensity due to the uniformly charged wire of infinite length:
Let us consider a uniformly-charged (positively charged) wire of infinite length having a constant linear charge density (that is, a charge per unit length) $\lambda$ coulomb/meter. Let P is a point at a distance $r$ from the wire at which electric field $\overrightarrow{E}$ has to find.
Let us draw a coaxial Gaussian cylindrical surface of length $l$ through point $P$. By symmetry, the magnitude $E$ of the electric field will be the same at all points on this surface and directed radially outward.
Thus, Now take small area elements $dA_{1}$, $dA_{2}$ and $dA_{3}$ on the Gaussian surface as shown in figure below. Therefore, the total electric flux passing through area elements is:
$ \phi_{E}= \oint \overrightarrow{E} \cdot \overrightarrow{dA_{1}} + \oint \overrightarrow{E} \cdot \overrightarrow{dA_{2}} + \oint\overrightarrow{E} \cdot \overrightarrow{dA_{3}}$
$ \phi_{E}= \oint E\: dA_{1} cos\theta_{1} + \oint E\: dA_{2} cos\theta_{2} \\ \qquad + \oint E\: dA_{3} cos\theta_{3}$
The angle between the area element $dA_{1}$, $dA_{2}$ and $dA_{3}$ with electric field are $\theta_{1}= 0^{\circ}$ , $\theta_{2}= 90^{\circ}$, and $\theta_{3}=90^{\circ}$ respectively. So total electric flux
$ \phi_{E}= \oint E\: dA_{1} cos0^{\circ} + \oint E\: dA_{2} cos90^{\circ} \\ \qquad + \oint E\: dA_{3} cos90^{\circ}$
Here $cos \: 0^{\circ}=1$ and $cos \: 90^{\circ}=0$
Hence the above equation can be written as:
$ \phi_{E}= \oint E\:dA_{1} $
The total electric flux passing through the Gaussian surface is
$ \phi_{E}= \oint{E\:dA_{1}} $
$ \phi_{E}= E\:\oint{dA_{1}} $
$ \phi_{E}= E\:\left(2\pi r l \right) \qquad \left\{\because \oint{dA_{1}} =2\pi r l \right\} $
$ \phi_{E}= E\:\left(2\pi r l \right)$
But, By Gaussian's law, The total flux $\phi_{E}$ must be equal to $\frac{q}{\epsilon_{0}}$, where $q$ is the total charge enclosed with the Gaussian surface. so that
$ \frac{q}{\epsilon_{0}}= E\:\left(2\pi r l \right)$
$ E= \frac{q}{2\pi r l \epsilon_{0}}$
For linear charge distribution →
$q=\lambda l$
So substitute this value in the above equation which can be written as
$ E= \frac{\lambda l}{2\pi r l \epsilon_{0}}$
$ E= \frac{\lambda}{2\pi\epsilon_{0} r }$
The vector form of the above equation :
$\overrightarrow{E}= \frac{\lambda}{2\pi\epsilon_{0} r }\widehat{r}$
Where $\widehat{r}$ is a unit vector in the direction of $r$. The direction of $\overrightarrow{E}$ is radially outwards(for positively charged
wire).
Thus, the electric field ($E$) due to the linear charge is inversely proportional to the distance ($r$) from the linear charge and its direction
is outward perpendicular to the linear charge.
Special Note:
A charged cylindrical conductor behaves for external points as the whole charge is distributed along its axis.
Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig
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