Dielectric

Substances having polar atoms/molecules intrinsically or being polarized are called dielectrics.

Polar Dielectric

Substances which have polar atoms/molecules intrinsically but are randomly arranged. On application of external electric field they get polarized parallel to the external electric field.

Net electric field inside dielectric.

$\vec{E}={\vec{E}}_0-{\vec{E}}_i$ ${\vec{E}}_i$= electric field due to induced charges

${\vec{E}}_i={\vec{E}}_0\left[1-\frac{1}{K}\right]$ $\sigma$ – surface charge density of capacitor plates

$q_i=q\left[1-\frac{1}{K}\right]$ ${\sigma }_i$ – induced charge density

${\sigma }_i=\sigma \left[1-\frac{1}{K}\right]$

Capacitance for different geometrical figures

(i) Parallel plate capacitor

C = $\frac{{\in }_{0}KA}{d}$

A- area of each plate,

d - distance between plates,

K-dielectric constant of the medium between the plates.

C = $\frac{{\in }_{0}A}{d}$

(ii) When a number of mediums are placed between the plates-

$C=\frac{{\in }_{0}A}{\left(\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}\right)}$

where t₁ + t₂ + t₃ = d

(iii) In this combination equivalent capacitance

C = $\frac{{\text{K}}_{\text{1}}{\in }_0{A}_1}{d}+\frac{{\text{K}}_{\text{2}}{\in }_0{A}_2}{d}+\frac{{\text{K}}_{\text{3}}{\in }_0{A}_3}{d}$

C = $\frac{{\in }_0}{d}$ (K₁A₁ + K₂A₂ + K₃A₃)

(iv) If a metal sheet of thickness 't' is kept between the two plates, then C = $\frac{{\in }_{0}A}{d-t}$.

(v) If a thin metal foil is placed between the plates, then the capacity remains uneffected.

(vi) $\frac{C_{medium}}{C_{air}}$= K

Note(1) :A parallel plate capacitor is connected to a battery. What will happen when

Case-I Distance between the plates is reduced-

(i) Same potential difference

(ii) Increased capacitance

(iii) Charge will increase

(iv) Electric field will increase

(v) Energy stored will increase.

Case-II Dielectric is kept between the plates-

(i) Same potential difference

(ii) Capacitance will increase

(iii) Charge will increase

(iv) No change in electric field

(v) Energy stored will increase, this increase is a result of work done to place the dielectric between the plates.

Note (2):-A parallel plate capacitor is charged and then battery is removed, when -

Case-I Distance between the plates is decreased-

(a) Charge will remains the same

(b) Capacitance will increase

(c) Potential difference across plates will decrease

(d) No change in electric field.

(e) Energy stored will decrease. This decrease in energy is due to work done in bringing the plates closer.

Case -II A dielectric is placed between the plates-

(a) No change in charge

(b) Increase in capacitance

(c) Potential difference will decrease

(d) Electric field between the plates will decreases.

(e) Energy stored will be reduced.

Note:(i) If nothing is mentioned then assume battery is disconnected and Q is constant.

(ii) A parallel plate capacitor is connected to battery (V-constant) and a slab of dielectric constant K is inserted between the plates then total energy given by battery is divided into two parts.

(a) Half is used to insert the slab

(b) Half is stored in form of E.P.E.

Spherical Condenser

(1) When outer sphere is earthed

C = 4$\pi {\in }_0$ $\frac{R_1{R}_2}{(R_2-R_1)}$

(2) When inner sphere is earthed

This situation can be visualized as a parallel combinations of two capacitors. One between the two spheres and second between outer sphere and infinity-

$\therefore$ C = $\frac{4\pi {\in }_0{K}_1{R}_1{R}_2}{(R_2-R_1)}$ + 4$\pi {\in }_0$K₂R₂

if K₁ = K₂ = K then C = $\frac{4\pi {\in }_{0}K{R}_2^2}{(R_2-R_1)}$

If 'Q' charge is given to outer sphere.

It is divided partly between it's inner and outer surface.

Let the charge on inner spheres be 'q', potential on inner sphere = 0

So charge on inner surface of outer sphere

= q = $\frac{R_1}{R_2}$Q

Charge on outer sphere

= (Q – q) = $\frac{R_2-R_1}{R_2}$Q

Charge on inner sphere = q = – $\frac{R_1}{R_2}$Q

Capacitance of cylindrical capacitor

If R1 : Radius of inner

R2 : Radius of outer cylinder

$\ell$ : length of cylinder then,

C = $\frac{2\pi {\in }_{0}K\ell }{\ln \frac{R_2}{R_1}}$= $\frac{2\pi {\in }_{0}K\ell }{2.303{\log }_{10}\left(\frac{R_2}{R_1}\right)}$

(1) Connected Spheres

C = 4$\pi {\in }_0$R₂

(2) Connected spheres (far away)

C = C₁ + C₂

C = 4$\pi {\in }_0$ (R₁ + R₂)