- Introduction
- |
- Calculation of capacitance
- |
- Parallel plate capacitor
- |
- Cylinderical capacitor
- |
- Spherical capacitor
- |
- Capacitors in series and parallel combinations
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- Energy stored in a capacitor
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- Effect of Dielectric

- Figure below shows two capacitors connected in parallel between two points A and B

- Right hand side plate of capacitors would be at same common potential V
_{A}. Similarly left hand side plates of capacitors would also be at same common potential V_{B}.

- Thus in this case potential difference V
_{AB}=V_{A}-V_{B}would be same for both the capacitors, and charges Q_{1}and Q_{2}on both the capacitors are not necessarily equal. So,

Q_{1}=C_{1}V and Q_{2}=C_{2}V

- Thus charge stored is divided amongst both the capacitors in direct proportion to their capacitance.

- Total charge on both the capacitors is,

Q=Q_{1}+Q_{2}

=V(C_{1}+C_{2})

and

Q/V=C_{1}+C_{2}(8)

So system is equivalent to a single capacitor of capacitance

C=Q/V

where,

- When capacitors are connected in parallel their resultant capacitance C is the sum of their individual capacitances.

- The value of equivalent capacitance of system is greater then the greatest individual one.

- If there are number of capacitors connected in parallel then their equivalent capacitance would be

C=C_{1}+C_{2}+ C_{3}........... (10)

**(ii) Series combination of capacitors**

- Figure 7 below shows two capacitors connected in series combination between points A and B.

- Both the points A and B are maintained at constant potential difference V
_{AB}.

- In series combination of capacitors right hand plate of first capacitor is connected to left hand plate of next capacitor and combination may be extended foe any number of capacitors.

- In series combination of capacitors all the capacitors would have same charge.

- Now potential difference across individual capacitors are given by

V_{AR}=Q/C_{1}

and,

V_{RB}=Q/C_{2}

- Sum of V
_{AR}and V_{RB}would be equal to applied potential difference V so,

V=V_{AB}=V_{AR}+V_{RB}

=Q(1/C_{1}+ 1/C_{2})

or,

where

i.e., resultant capacitance of series combination C=Q/V, is the ratio of charge to total potential difference across the two capacitors connected in series.

- So, from equation 12 we say that to find resultant capacitance of capacitors connected in series, we need to add reciprocals of their individual capacitances and C is always less then the smallest individual capacitance.

- Result in equation 12 can be summarized for any number of capacitors i.e.,

For prectical applications , two or more capacitors are often used in combination and their total capacitance C must be known.To find total capacitance of the arrangement of capacitor we would use equation

Q=CV

- Consider a capacitor of capacitance C, completely uncharged in the begning.

- Charhing process of capacitor requires expanditure of energy because while charging a capacitor charge is transferred from plate at lower potential to plate at higher potential.

- Now if we start charging capacitor by transporting a charge dQ from negative plate ti the positive plate then work is done against the potential difference across the plate.

- If q is the amount of charge on the capacitor at any stage of charging process and φ is the potential difference across the plates of capacitor then magnitude of potential difference is φ=q/C.

- Now work dW required to transfer dq is

dW=φdq=qdq/C

- To charge the capacitor starting from the uncharged state to some final charge Q work required is

Integrating from 0 to Q

W=(1/C)∫qdq

=(Q^{2})/2C (14a)

=(CV^{2})/2

=QV/2

Which is the energy stored in the capacitor and can also be written as

U=(CV^{2})/2 ---(15)

- From equation 14c,we see that the total work done is equal to the average potential V/2 during the charging process ,multiplied by the total charge transferred

- If C is measured in Farads ,Q in coulumbs and V in volts the energy stored would in Joules

- A parallel plate capacitor of area A and seperation d has capacitance

C=ε_{0}A/d

- electric field in the space between the plates is

E=V/d or V=Ed

Putting above values of V and C in equation 14b we find

W=U=(1/2)(ε_{0}A/d)(Ed)^{2}

=(1/2)ε_{0}E^{2}(Ad)

=(1/2)ε_{0}E^{2}.V ---(16)

- If u denotes the energy per unit volume or energy density then

u=(1/2)ε_{0}E^{2}x volume

- The result for above equation is generally valid even for electrostatic field that is not constant in space.

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

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