The Potential Energy of a system of two-point like charges→
When the system of two charged particles is configured, in which one charge
is at rest of position and another is brought from infinity to near the first charge then the work done acquire by this charged particle is stored
in the form of electric potential energy between these charges.
Derivation→
Let us consider, If two charge $q_{1}$ and $q_{2}$ in which one charge $q_{1}$ is at the rest of the position at point $P_{1}$ and another charge
$q_{2}$ is brought from infinity to a point $P_{2}$ to configure the system then the electric potential at point $P_{2}$ due to charge particle $q_{1}$ →
$V=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}}{r}$
Where $r$ is the distance between the point $P_{1}$ and Point $P_{2}$
Here, Charge $q_{2}$ is moved in from infinity to point $P_{2}$ then the work required is →
$W=V q_{2}$
$W= \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r}$
Since the electric potential at infinity is zero the work- done will also be zero. So total work-done from infinity to a point $P$ will be stored in the form of electric potential energy.
$U=W$
$ U= \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r}$
The electric potential energy of a system of three-point-like charges→
To obtain the potential energy of a system of three charges. First, Obtain the work done between any two charges and then obtain the different work done for both those charges from the third charge, and then the total work done will be equal to electric potential energy.
Let us consider a system is made up of three charges $q_{1}$, $q_{2}$ and $q_{3}$ which are placed at point $P_{1}$,$P_{2}$ and $P_{3}$. Now the work done between two charges $q_{1}$ and $q_{2}$ is
$W_{1}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r_{12}}$
Now, the charge $q_{3}$ is brought from infinity to the point $P_{3}$. Work has to be done against the forces exerted by $q_{1}$ and $q_{2}$
Therefore, The work done between charges of $q_{2}$ and $q_{3}$
$W_{2}=\frac{1}{4\pi\epsilon_{0}} \frac{q_{2}q_{3}}{r_{23}}$
Now, The work done between charges $q_{1}$ and $q_{3}$
$ W_{3}=\frac{1}{4\pi\epsilon_{0}} \frac{q_{1}q_{3}}{r_{13}}$
The total work done to make a system of three charges:
$ W=W_{1}+W_{2}++W_{3}$
Now substitute the value of $W_{1}$, $W_{2}$and $W_{2}$ in above equation i.e.
$ W=\frac{1}{4\pi\epsilon_{0}} \left[\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\right]$
This work is stored in the form of electric potential energy in the system.
$U=W$
$ U=\frac{1}{4\pi\epsilon_{0}} \left[\frac{q_{1}q_{2}}{r_{12}}+ \frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\right]$
Similarly, the Potential energy of a system of N point system i.e.
$ U=\frac{1}{4\pi\epsilon_{0}} \sum_{i=1}^{N}\sum_{j=1}^{N}\frac{q_{i}q_{j}}{r_{ij}}$
Here $i\neq j$
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
Comments
Post a Comment
If you have any doubt. Please let me know.