A capacitor is usually charged by connecting it to a battery. To charge up the capacitor, we take a small elemental positive charge "+dq" from a neutral system and move it a distance "d" so, that the system gets "-dq". The work is done against the Electric field which directs opposite to the motion.
Let say, the amount of charge piled in the positive plate is +q, so that the potential difference between the plates is "q/C".
Now, the work required to do move the next "dq" charge is simply just the potential multiplied by the amount of charge [The definition of potential is work done per unit charge].
So, $$ dW = \frac{q}{C} dq $$
Then the total work done to charge up the plate to +Q charge is, $$ W = \int_0^Q \frac{q}{C} dq = \frac{Q^2}{2C} $$ Using the relation, $ Q = CV $ we get, $$ W = \frac{1}{2} CV^2 $$ which is the electrostatic potential energy stored in the system.
For example, in the case of parallel plate capacitor, the energy stored can be calculated as, where $ C = \frac{\epsilon_0 A}{d} $ and V = Ed ,$$ W = \frac{CV^2}{2} = \frac{\epsilon_0 A E^2 d }{2} $$
Here, "Ad" represents the volume of the in between region where the energy is stored. So, we can define a new term as,
Electrostatic Energy density = $\frac{1}{2} \epsilon_0 E^2 $
As an example, if we try to calculate the electrostatic energy density of air at the break down voltage which is about $ E = 3\times 10^6 Vm^{-1} $ and the energy density can be calculated as, 39.825 $Jm^{-3} $, the value is so high.
One Joule is defined as the work required to move one kilogram object to one meter distance. It is nearly 40 Joules of energy in a volume one meter cube.
Now, we will give a slight attention to dielectric system. We know there are two types of charges are produced in polarization. First one is the bound charge, $ \rho_b = - \nabla\cdot\vec{P} $ and the surface charge, $ \sigma_b = \vec{P}\cdot \hat{n} $
To apply Gauss law, the total charge inside the material is identified as the free charge and the bound charge. $$ \rho _{total}= \rho_{bound} + \rho_{free} $$
Using gauss law, $$ \rho_{total} = \epsilon_0\nabla\cdot\vec{E} = -\nabla\cdot\vec{P} +\rho_{free} $$
Calling, $$ \vec{D}= \epsilon_0\vec{E} + \vec{P} $$ we get, $$ \nabla\cdot\vec{D} = \rho_{free} $$
In most of the cases, when Electric field is not so much high, Polarization is given by, $$ \vec{P} = \epsilon_0 \chi_e \vec{E} $$
where $ \chi_e $ is called the electric susceptibility.
Then, $$ \vec{D} = \epsilon_0 (1+\chi_e) \vec{E}$$ $$\vec{D}= \epsilon \vec{E} $$ and $$ \epsilon_r = \frac{\epsilon}{\epsilon_0}$$ is called relative permitivity or dielectric constant.
For vacuum, $$ \vec{D} = \epsilon_0 \vec{E_{ext}} $$ since there is no polarization to take place. In a dielectric medium $$ \vec{D} = \epsilon \vec{E_{in}} = \epsilon_r\epsilon_0 \vec{E_{in}} = \epsilon_0 \vec{E_{ext}} $$
Thus it gives, $$ \vec{E_{ext}} = \epsilon_r \vec{E_{in}} \rightarrow \vec{E_{in}} = \frac{\vec{E_{ext}}}{\epsilon_r} $$ which says that the Electric field is reduced to "$\frac{1}{\epsilon_r} $" factors than the Electric field applied (in the vacuum).
Then, we can see that, if a dielectric medium is introduced in the between region of capacitor plates, capacitance becomes $$ C'= \frac{Q'}{V'} = \frac{Q}{V'}$$ since the charge is the same in both the cases. But, $$ V' = \frac{E'}{d} = \frac{E}{\epsilon_r{d}} $$ and $$ C' = \epsilon_r C_{vacuum} $$
which means the capacitance of the capacitor is increased by the factor of dielectric constant value of the inserted dielectric medium.
Using this, we can find the expression for the new capacitance of a parallel plate capacitor where dielectric material of thickness "t" is introduced.
To find the new potential difference, $$ V' = - \int_-^+ \vec{E}\cdot \vec{dl} = \int_0^{d-t}E_{vacuum} dl + \int_0^t E_{dielectric} dl $$
which gives, $$ V' = E_{vacuum}(d-t) + \frac{E_{vacuum}}{\epsilon_r} t $$
Electric field between the plates is given by, $$ E_{vacuum} = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0} $$
So, $$ V' = \frac{Q}{A\epsilon_0} [ (d-t) + \frac{t}{\epsilon_r}] = \frac{Q}{A\epsilon_0} [d-t(1-\frac{1}{\epsilon_r})]$$
And the capacitance is given by, $$ C' = \frac{Q}{V'} = \frac{A\epsilon_0}{d-t(1-\frac{1}{\epsilon_r})} $$
Let say, the amount of charge piled in the positive plate is +q, so that the potential difference between the plates is "q/C".
Now, the work required to do move the next "dq" charge is simply just the potential multiplied by the amount of charge [The definition of potential is work done per unit charge].
So, $$ dW = \frac{q}{C} dq $$
Then the total work done to charge up the plate to +Q charge is, $$ W = \int_0^Q \frac{q}{C} dq = \frac{Q^2}{2C} $$ Using the relation, $ Q = CV $ we get, $$ W = \frac{1}{2} CV^2 $$ which is the electrostatic potential energy stored in the system.
For example, in the case of parallel plate capacitor, the energy stored can be calculated as, where $ C = \frac{\epsilon_0 A}{d} $ and V = Ed ,$$ W = \frac{CV^2}{2} = \frac{\epsilon_0 A E^2 d }{2} $$
Here, "Ad" represents the volume of the in between region where the energy is stored. So, we can define a new term as,
Electrostatic Energy density = $\frac{1}{2} \epsilon_0 E^2 $
As an example, if we try to calculate the electrostatic energy density of air at the break down voltage which is about $ E = 3\times 10^6 Vm^{-1} $ and the energy density can be calculated as, 39.825 $Jm^{-3} $, the value is so high.
One Joule is defined as the work required to move one kilogram object to one meter distance. It is nearly 40 Joules of energy in a volume one meter cube.
Now, we will give a slight attention to dielectric system. We know there are two types of charges are produced in polarization. First one is the bound charge, $ \rho_b = - \nabla\cdot\vec{P} $ and the surface charge, $ \sigma_b = \vec{P}\cdot \hat{n} $
To apply Gauss law, the total charge inside the material is identified as the free charge and the bound charge. $$ \rho _{total}= \rho_{bound} + \rho_{free} $$
Using gauss law, $$ \rho_{total} = \epsilon_0\nabla\cdot\vec{E} = -\nabla\cdot\vec{P} +\rho_{free} $$
Calling, $$ \vec{D}= \epsilon_0\vec{E} + \vec{P} $$ we get, $$ \nabla\cdot\vec{D} = \rho_{free} $$
In most of the cases, when Electric field is not so much high, Polarization is given by, $$ \vec{P} = \epsilon_0 \chi_e \vec{E} $$
where $ \chi_e $ is called the electric susceptibility.
Then, $$ \vec{D} = \epsilon_0 (1+\chi_e) \vec{E}$$ $$\vec{D}= \epsilon \vec{E} $$ and $$ \epsilon_r = \frac{\epsilon}{\epsilon_0}$$ is called relative permitivity or dielectric constant.
For vacuum, $$ \vec{D} = \epsilon_0 \vec{E_{ext}} $$ since there is no polarization to take place. In a dielectric medium $$ \vec{D} = \epsilon \vec{E_{in}} = \epsilon_r\epsilon_0 \vec{E_{in}} = \epsilon_0 \vec{E_{ext}} $$
Thus it gives, $$ \vec{E_{ext}} = \epsilon_r \vec{E_{in}} \rightarrow \vec{E_{in}} = \frac{\vec{E_{ext}}}{\epsilon_r} $$ which says that the Electric field is reduced to "$\frac{1}{\epsilon_r} $" factors than the Electric field applied (in the vacuum).
Then, we can see that, if a dielectric medium is introduced in the between region of capacitor plates, capacitance becomes $$ C'= \frac{Q'}{V'} = \frac{Q}{V'}$$ since the charge is the same in both the cases. But, $$ V' = \frac{E'}{d} = \frac{E}{\epsilon_r{d}} $$ and $$ C' = \epsilon_r C_{vacuum} $$
which means the capacitance of the capacitor is increased by the factor of dielectric constant value of the inserted dielectric medium.
Using this, we can find the expression for the new capacitance of a parallel plate capacitor where dielectric material of thickness "t" is introduced.
To find the new potential difference, $$ V' = - \int_-^+ \vec{E}\cdot \vec{dl} = \int_0^{d-t}E_{vacuum} dl + \int_0^t E_{dielectric} dl $$
which gives, $$ V' = E_{vacuum}(d-t) + \frac{E_{vacuum}}{\epsilon_r} t $$
Electric field between the plates is given by, $$ E_{vacuum} = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0} $$
So, $$ V' = \frac{Q}{A\epsilon_0} [ (d-t) + \frac{t}{\epsilon_r}] = \frac{Q}{A\epsilon_0} [d-t(1-\frac{1}{\epsilon_r})]$$
And the capacitance is given by, $$ C' = \frac{Q}{V'} = \frac{A\epsilon_0}{d-t(1-\frac{1}{\epsilon_r})} $$
No comments:
Post a Comment
Let everyone know what you think about this