Physics > Capacitors > 9.0 Dielectrics (Insulators) and Polarization
Capacitors
1.0 Introduction
2.0 Different types of capacitors and its capacitance.
3.0 Parallel Plate Capacitor
3.1 When equal and opposite charges placed on plates
3.2 When unequal charges are placed on the two plates
4.0 Capacitance of spherical conductor
5.0 Capacitance of a earthed sphere by a concentric spherical shell
6.0 Capacitance of a cylindrical capacitor
7.0 Mechanical force on the charged conductor
8.0 Redistribution of Charge
9.0 Dielectrics (Insulators) and Polarization
9.1 Effect of Dielectric
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)
10.0 Combination of capacitors
11.0 Energy Density ($u$)
12.0 $R$-$C$ Circuits
13.0 Method of Finding Equivalent Capacitance
14.0 Some important concepts
15.0 Van De Graaff Generator
9.1 Effect of Dielectric
3.2 When unequal charges are placed on the two plates
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)
Placing a solid dielectric between the plates of a capacitor solves the following three problem:
1. It solves the problem of maintaining two large metal sheets at a very small separation without actual contact.
2. It increases the maximum possible potential difference which can be applied between the plates of the capacitor without the dielectric breakdown. A lot of dielectric material can bear a stronger electric without breakdown field then air.
3. It increases the capacity of the capacitor.
When a dielectric material is inserted between the plates (keeping charge to be constant) of a capacitor, the electric
field and hence the potential difference decreases by a factor $K$ (the dielectric constant of the dielectric).
field and hence the potential difference decreases by a factor $K$ (the dielectric constant of the dielectric).$$E = \frac{{{E_ \circ }}}{K}$$ and $$ V = \frac{{{V_ \circ }}}{K}$$(when $q$ is constant)
Electric field decreases because an induced charge of opposite sign induce on each surface of the dielectric. This induced charge generate an electric field inside the dielectric in opposite direction and the resulting electric field is less than the original one. We can find the induced charge as following:
$$E = {E_ \circ } - {E_i}$$or $$\frac{{{E_ \circ }}}{K} = {E_ \circ } - {E_i}$$
$${E_i} = {E_ \circ }\left( {1 - \frac{1}{K}} \right)$$
$$\frac{{{\sigma _i}}}{{{\varepsilon _ \circ }}} = \frac{\sigma }{{{\varepsilon _ \circ }}}\left( {1 - \frac{1}{K}} \right)$$
$${\sigma _i} = \sigma \left( {1 - \frac{1}{K}} \right)$$
$${q_i} = q\left( {1 - \frac{1}{K}} \right)$$
If $K=0$ then, $${q_i} = q,{\sigma _i} = \sigma$$ and $$E = {E_ \circ }$$
else $${q_i} < q$$
Now we can conclude our discussion about electric field in presence of dielectric, like this :
1. ${E_{vacuum}} = {E_ \circ } = \frac{\sigma }{{{\varepsilon _ \circ }}} = \frac{q}{{A{\varepsilon _ \circ }}}$
2. ${E_{dielectric}} = \frac{{{E_ \circ }}}{K}$
3. ${E_{conductor}} = 0$ (because $K = 0$)
If we plot a graph between potential and distance from positive plate.
The modulus of slope of $AB$ = slope of $CD$ = slope of $EF$ = ${E_\circ}$
Slope of $BC$ = $\frac{{{E_ \circ }}}{K}$ and slope of $DE=0$
Now, the potential difference between the plates is $$\begin{equation} \begin{aligned} {V_ + } - {V_ - } = {E_ \circ }d + \frac{{{E_ \circ }}}{K}.d + {E_ \circ }d + 0 + {E_ \circ }d \\ = 3{E_ \circ } + \frac{{{E_ \circ }}}{K}.d \\\end{aligned} \end{equation} $$ Here we used potential difference $E.d$)
