1. Capacitor & capacitance:

(i) A capacitor consists of two conductors carrying charges of equal magnitude and opposite sign.

(ii) The capacitance $C$ of any capacitor is the ratio of the charge $Q$ on either conductor to the potential difference $V$ between them $C = \frac{Q}{V}$ .

(iii) The capacitance depends only on the geometry of the conductors and not on an external source of charge or potential difference.

2. Dielectric:

Dielectrics are insulating (non-conducting) materials that transmit electric effects without conducting.

Dielectrics are of two types.

3. Polar dielectrics:

A polar molecule has a permanent electric dipole moment $(ρ)$ in the absence of an electric field also. But a polar dielectric has a net dipole moment zero in the absence of an electric field because polar molecules are randomly oriented as shown in the figure.

4. Nonpolar dielectric:

(i) In non-polar molecules, each molecule has zero dipole moment in its normal state.

(ii) When an electric field is applied, molecules become induced electric dipoles e.g., $N_{2}, O_{2}$ , Benzene, Methane, etc. are made of non-polar atoms/molecules.

5. Dielectric breakdown and dielectric strength:

(i) If a very high electric field is created in a dielectric, the dielectric then behaves like a conductor. This phenomenon is known as a dielectric breakdown.

(ii) The maximum value of electric field (or potential gradient) that a dielectric material can tolerate without its electric breakdown is called its dielectric strength.

(iii) S.I. unit of dielectric strength of a material is $\frac{V}{m}$ but the practical unit is $\frac{kV}{mm}$ .

6. Parallel plate capacitor:

(i) $q \propto V \Rightarrow q = C V$ , where, $q =$ charge on positive plate of the capacitor, $C :$ Capacitance of capacitor and, $V :$ Potential difference between positive and negative plates.

(ii) Capacitance, $C = \frac{ε_{0} A}{d}$ , where $A$ is surface area of plate and d is the distance between the plates.

(iii) When complete space between the plates is occupied by a medium of dielectric constant K, then $C' = \frac{K ε_{0} A}{d} = K C$ .

(iv) Electric field between the plates: $E = \frac{σ}{ε_{0}} = \frac{Q}{A ε_{0}}$

(v) Energy density between the plates of the capacitor, $u = \frac{Energy}{Volume} = \frac{1}{2} ε_{0} E^{2} .$

(vi) Representation of capacitor:

(vii) Attractive force between the capacitor plates, $F = (\frac{σ}{2 ε_{0}}) (σ A) = \frac{Q^{2}}{2 ε_{0} A}$

7. Capacitor with dielectric:

(i) Capacitance in the presence of dielectric:

(a) $C = \frac{K ε_{0} A}{d} = K C_{0}$

$C_{0} =$ Capacitance in the absence of dielectric.

(b) $E_{net} = E - E_{ind} = \frac{σ}{ε_{0}} - \frac{σ_{b}}{ε_{0}} = \frac{σ}{K ε_{0}} = \frac{V}{d}$

(d) Field inside the slab is $E_{n e t} = \frac{E}{K}$

(e) $E_{ind} = E (1 - \frac{1}{K})$

(f) Induced (bound) charge density.

$σ_{ind} = σ (1 - \frac{1}{K})$

8. Capacitance with different dielectrics:

(i) If a dielectric slab is partially filled between the plates

$\Rightarrow C' = \frac{ε_{0} A}{d - t + \frac{t}{K}}$

Note: for metal plate inside the capacitor, $K = \infty$ . So, $C' = \frac{ε_{0} A}{d - t}$

(ii) Capacitor with multiple dielectrics:

$C' = \frac{ε_{0} A}{[d - (t_{1} + t_{2} + t_{3} \dots . . + t_{n}) + (\frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}} + \frac{t_{3}}{K_{3}} + \dots \dots . . + \frac{t_{n}}{K_{n}})]}$

(iii) Capacitor with three different dielectrics:

(a) Case I:

$C_{1} = \frac{K_{1} ε_{0} A_{1}}{d_{1} + d_{2}}$ , $C_{2} = \frac{K_{2} ε_{0} A_{2}}{d_{1}}$ and $C_{3} = \frac{K_{3} ε_{0} A_{2}}{d_{2}}$

The effective capacitance of the combination is, $C_{e f f} = C_{1} + \frac{C_{2} C_{3}}{C_{2} + C_{3}}$ .

(b) Case II:

$C_{1} = \frac{K_{1} ε_{0} A_{1}}{d_{1}}$ , $C_{2} = \frac{K_{2} ε_{0} A_{2}}{d_{1}}$ and $C_{3} = \frac{K_{3} ε_{0} (A_{1} + A_{2})}{d_{2}}$

The effective capacitance is, $C_{e f f} = \frac{C_{3} (C_{1} + C_{2})}{C_{1} + C_{2} + C_{3}}$

9. Capacitance of an Isolated Spherical Conductor:

(i) $C = 4 π ε_{0} ε_{r} R$ in a medium

(ii) $C = 4 π ε_{0} R$ in air

10. Spherical Capacitor:

It consists of two concentric spherical shells as shown in the figure. Here capacitance of the region between the two shells is $C_{1}$ and that outside the shell is $C_{2}$ . We have

$C_{1} = \frac{4 π ε_{0} a b}{b - a}$ and $C_{2} = 4 π ε_{0} b$

(i) Case I: If the outer shell is earthed, then the capacitance

$C = C_{1} = \frac{4 π ε_{0} a b}{b - a}$

(ii) Case II: If the inner shell is earthed, then $C_{1}$ and $C_{2}$ will be in parallel.

So, the effective capacitance becomes $C = C_{1} + C_{2} = \frac{4 π ε_{0} a b}{b - a} + 4 π ε_{0} b$

(iii) Case III: If there exists a potential difference between inner shell and infinity, then $C_{1}$ and $C_{2}$ will be in series.

$\frac{1}{c_{e f f}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{b - a}{4 π ε_{0} a b} + \frac{1}{4 π ε_{0} b} = 4 π ε_{0} a$

11. Cylindrical capacitor: It consists of two concentric cylinders of radii $a$ and $b$ $(a < b)$ , inner cylinder is given charge +Q while the outer cylinder is earthed. The common length of the cylinders is $l$ then

$C = \frac{2 π ε_{0} l}{\log_{e} (\frac{b}{a})}$

12. Energy stored in the capacitor:

(i) $U = \frac{1}{2} C V^{2} = \frac{Q^{2}}{2 C} = \frac{Q V}{2}$

(ii) Energy density $= \frac{1}{2} ε_{0} ε_{r} E^{2} = \frac{1}{2} ε_{0} K E^{2}$ . Where, $ε_{r} =$ Relative permittivity of the medium, $K = ε_{r} :$ dielectric Constant.

(iii) For vacuum, energy density $= \frac{1}{2} ε_{0} E^{2}$

(iv) In charging a capacitor by battery half the energy supplied is stored in the capacitor and the remaining half energy $(\frac{1}{2} Q V)$ is lost in the form of heat.

13. Effect of dielectric on the energy stored in a capacitor:

(i) Case I: Keeping the battery connection if a dielectric is inserted between the plates of a capacitor such that the gap is completely occupied, then the energy becomes $U_{f} = K U_{i}$

(ii) Case II: Keeping the charge constant if a dielectric is inserted between the plates of a capacitor such that the gap is completely occupied, then the energy becomes,

$U_{f} = \frac{U_{i}}{K}$

14. Heat Produced in a capacitive circuit:

Heat $=$ Work done by battery $-$ change in potential energy of capacitors.

(i) Case 1:

In the given figure, the heat lost on reversing the terminals of the battery is,

$H = 2 C E^{2}$

(ii) Case 2:

In the given figure, the heat lost when the switch is shifted from 1 to 2 is

$H = \frac{C C_{0} E^{2}}{2 C + C_{0}}$

15. Combination of capacitors:

(i) Series Combination:

(a) Charge on all capacitors is the same.

(b) $\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

(ii) Parallel Combination:

(a) Potential difference is the same across all capacitors.

(b) $C_{eq} = C_{1} + C_{2} + C_{3}$

16. Coalesce of charged drops:

Suppose we have $n$ identical drops each having radius $r$ , capacitance $c$ , charge $q$ , potential $v$ and potential energy $u$ .

If these drops are combined to form a big drop of radius $R$ , capacitance $C$ , charge $Q$ , potential $V$ and potential energy $U$ , then

(i) Charge on big drop: $Q = n q$

(ii) Radius of big drop: Volume of big drop $= n$ volume of a single drop

i.e., $\frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3}$ , $R = n^{1 / 3} r$

(iii) Capacitance of big drop: $C = n^{1 / 3} c$

(iv) Potential of big drop: $V = \frac{Q}{C} = \frac{n q}{n^{\frac{1}{3}} c}$ $V = n^{\frac{2}{3}} v$

(v) Energy of big drop: $U = \frac{1}{2} C V^{2} = \frac{1}{2} (n^{\frac{1}{3}} c) {(n^{\frac{2}{3}} v)}^{2} v$ $\Rightarrow U = n^{\frac{5}{3}} u$

(vi) Energy difference: Total energy of big drop is greater than the total energy of all smaller drops. Hence energy difference $Δ U = U - n u = U - n \times \frac{U}{n^{\frac{5}{3}}} = U (1 - \frac{1}{n^{\frac{2}{3}}})$

17. Distribution of charges on connecting two charged capacitors:

When two capacitors are $C_{1}$ and $C_{2}$ are connected as shown in figure

(i) Common potential: $V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{T o t a l c h a r g e}{T o t a l c a p a c i t a n c e}$

(ii) $Q_{1}^{'} = C_{1} V = \frac{C_{1}}{C_{1} + C_{2}} (Q_{1} + Q_{2})$

(iii) $Q_{2}^{'} = C_{2} V = \frac{C_{2}}{C_{1} + C_{2}} (Q_{1} + Q_{2})$

(iv) Heat loss during redistribution:

$Δ H = U_{i} - U_{f} = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2}$ .

The loss of energy is in the form of Joule heating in the wire.

18. Force on dielectric:

(i) When the battery is connected $F = \frac{ε_{0} b (K - 1) V^{2}}{2 d}$

Note that the force is constant and independent of $x$ .

(ii) When the battery is not connected $F = \frac{Q^{2}}{2 C^{2}} \frac{d C}{d x}$

i.e., $F = \frac{q^{2} d (K - 1)}{2 ε_{0} b} {(\frac{1}{l + x (K - 1)})}^{2}$ . Where, $b$ is the width of the plate or dielectric slab.

(iii) Force on the dielectric will be zero when the dielectric is fully inside.

19. R-C circuit for DC source:

(i) Charging of Capacitor:

(a) Charge on Capacitor (Capacitor initially uncharged)

$q = q_{0} (1 - e^{\frac{- t}{τ}})$

$q_{0} =$ Charge on the capacitor at steady state

(b) $q_{0} = C V$

(d) $I = \frac{q_{0}}{τ} e^{\frac{- t}{τ}} = \frac{V}{R} e^{\frac{- t}{τ}}$

(ii) Discharging of Capacitor:

(a) $q = q_{0} e^{\frac{- t}{τ}}$

$q_{0} =$ Initial charge on the capacitor

(b) $I = \frac{q_{0}}{τ} e^{\frac{- t}{τ}}$

A parallel plate capacitor is charged and then isolated. On increasing the plate separation

Important Questions on Capacitance

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