Informative line

### Basic Concept

Learn basic concept and definition of capacitance, unit of capacitance. Practice formula to find capacitance of a parallel plate capacitor and electric field parallel plate capacitor, equivalent capacitance.

# Definition of Capacitance

• Consider conducting plates of a capacitor carrying charge of equal magnitude and opposite nature and potential difference between them is $$\Delta V$$.

• Experimentally, the relation between $$Q$$ and $$\Delta V$$ is given as,

$$Q\propto\Delta V$$

$$Q=C\Delta V$$

$$C=\dfrac{Q}{\Delta V}$$

• C is a constant known as capacitance of a capacitor and does not depend on Q or V.

## Unit of Capacitance

• Unit of capacitance is coulomb per volt or Farad.

$$1\,F=1\dfrac{C}{V}$$

• Farad is very large unit of capacitance. Generally devices have capacitances in microfarad and picofarad.

$$1\mu F=10^{-6}F$$

$$1p F=10^{-12}F$$

#### Consider a capacitor having charge Q on one of its conducting plate. What will be the change in capacitance is charge is doubled?

A Doubled

B Half

C Remains the same

D Becomes one-fourth

×

Capacitance (C) is a constant and does not depend on charge (Q) or voltage (V).

### Consider a capacitor having charge Q on one of its conducting plate. What will be the change in capacitance is charge is doubled?

A

Doubled

.

B

Half

C

Remains the same

D

Becomes one-fourth

Option C is Correct

# Capacitance of Parallel Plates Capacitor

• Consider two parallel metallic plates of equal area carrying a charge of equal magnitude and opposite nature are separated by a distance $$d$$.
• The surface charge density on each plate is $$\sigma$$

where $$\sigma=\dfrac{Q}{A}$$

• When plates are placed very closed to each other (compared to length and width) electric field is assumed to be uniform.

$$E=\dfrac{\sigma}{\epsilon_0} =\dfrac{Q}{\epsilon_0A}$$   $$\left[\sigma=\dfrac{Q}{A}\right]$$

$$\Delta V=E.d$$         from Gauss Law

$$\Delta V=\dfrac{Qd}{\epsilon_0A}$$

• Capacitance is given as $$C=\dfrac{Q}{\Delta V}$$

or, $$C=\dfrac{Q(\epsilon_0A)}{Qd}$$

or,  $$C=\dfrac{\epsilon_0A}{d}$$

• Capacitance depends on shape and size of plates but is independent of Q and V.

Conclusion

The capacitance of a parallel plate capacitor is proportional to the area of the plates and inversely proportional to the separation between the plates.

#### A parallel plate capacitor having area (A) of its plates are separated by a distance (d). What will be the capacitance (C) if area is doubled?

A Remains same

B Doubled

C Half

D Becomes one-fourth

×

The capacitance of capacitor is given as

$$C=\dfrac{\epsilon_0A}{d}$$

When,

$$A'=2A$$

New capacitance will be

$$C'=\dfrac{\epsilon_0(2A)}{d}=\dfrac{2\,\epsilon_0A}{d}$$

$$C'=2\,C$$

### A parallel plate capacitor having area (A) of its plates are separated by a distance (d). What will be the capacitance (C) if area is doubled?

A

Remains same

.

B

Doubled

C

Half

D

Becomes one-fourth

Option B is Correct

# Equivalent Capacitance due to Parallel Combination of Capacitors

• Two capacitors connected in parallel are joined from top to top and from bottom to bottom.
• Top electrode of both the capacitors are connected by a conducting wire so that they are at same potential.
• Similarly, bottom electrode of both the capacitors are connected by a conducting wire so that they are at same potential.
• Thus, it means two capacitors connected in parallel, have same potential difference.
• Consider a circuit in which two capacitors (C1 and C2) are connected in parallel with a battery.
• The potential difference across each capacitor is assumed to be $$\Delta V_c$$.
• Charge stored by capacitor C1

?$$Q_1=C_1\,\Delta V_c$$

• Charge stored by capacitor C2

$$Q_2=C_2\,\Delta V_c$$

• Now, the above circuit is replaced with new circuit having single capacitor in place of two capacitors having total charge $$Q=Q_1+Q_2$$

• The voltage across capacitor is $$\Delta V_c$$ .
• Thus, the capacitance of capacitor is given as

$$C_{eq}=\dfrac{Q}{\Delta V_c}$$

or,  $$C_{eq}=\dfrac{Q_1+Q_2}{\Delta V_c}$$

or,  $$C_{eq}=\dfrac{Q_1}{\Delta V_c}+\dfrac{Q_2}{\Delta V_c}$$

or,  $$C_{eq}=C_1+C_2$$

Conclusion :

If two or more capacitors are connected in parallel then their equivalent capacitance is given as

$$C_{eq}=C_1+C_2+C_3+...$$

#### Determine equivalent capacitance for the given circuit.

A $$3\,\mu F$$

B $$5\,\mu F$$

C $$7\,\mu F$$

D $$8\,\mu F$$

×

Equivalent capacitance for parallel combination is sum of capacitance of all capacitors.

Equivalent capacitance

$$C_{eq}=3\,\mu F+1\,\mu F+2\,\mu F+2\,\mu F$$

$$C_{eq}=8\,\mu F$$

### Determine equivalent capacitance for the given circuit.

A

$$3\,\mu F$$

.

B

$$5\,\mu F$$

C

$$7\,\mu F$$

D

$$8\,\mu F$$

Option D is Correct

# Calculation of Charge Provided by Battery

Case 1 :

• Consider a capacitor, initially having some charge on its plate of magnitude q.

• A battery with emf $$\mathcal E$$ is connected with the capacitor by throwing the switch from A to B.
• As battery is connected, battery charges the capacitor upto the maximum value i.e., $$C\mathcal E$$ .

$$Q_{max}=C\mathcal E$$  [$$\therefore$$Final charge on the capacitor after connection of battery]

• Charge provided by the battery

$$\Delta Q =Q_{max}-q$$      [$$\therefore$$ $$\Delta Q=Q_f-Q_i$$

$$\Delta Q=C\mathcal E-q$$

Case 2 :

Initially case configuration is as shown in figure.

• On moving switch from A to B, for positive plate charge provided by battery to charge upto maximum value i.e., $$\,C\mathcal E$$

$$\Delta Q = Q_{max} -(-q)$$

$$\Delta Q=C\mathcal E+q$$

Note- Charge provided by positive terminal of battery is  $$C\mathcal E+q$$  as battery first levels $$-q$$ to zero and then charges the plate upto $$C\mathcal E$$.

#### A parallel plate capacitor with capacitance $$C=7\,\mu F$$ having charge $$Q=3\,\mu C$$ on plate A is connected with a battery of  $$\mathcal E=1\,V.$$ Calculate the charge provided by the battery.

A $$4\,\mu C$$

B $$3\,\mu C$$

C $$8\,\mu C$$

D $$10\,\mu C$$

×

On moving switch from A to B, battery starts charging capacitor upto  $$Q_{max}$$.

Maximum charge on capacitor

$$Q_{max}=C\mathcal E$$

Given : $$C=7\,\mu F,\,\,\,\mathcal E=1\,V$$

$$Q_{max}=7\,\mu F\times 1\,V$$

$$Q_{max}=7\,\mu C$$

Charge provided by battery

$$\Delta Q=Q_{max}-Q$$

$$\Delta Q=7\,\mu C-3\,\mu C$$

$$\Delta Q=4\,\mu C$$

### A parallel plate capacitor with capacitance $$C=7\,\mu F$$ having charge $$Q=3\,\mu C$$ on plate A is connected with a battery of  $$\mathcal E=1\,V.$$ Calculate the charge provided by the battery.

A

$$4\,\mu C$$

.

B

$$3\,\mu C$$

C

$$8\,\mu C$$

D

$$10\,\mu C$$

Option A is Correct

# Charge across Capacitor Plates

• Consider a parallel plate capacitor connected across battery terminals.
• The final charge Q is independent of initial charge on the plates.
• Charge moves on the plates untill potential difference between plates becomes equal to battery's  $$e.m.f.$$

• After some time,

Potential difference across the plates = Potential difference across the terminals of battery.

• Charge on plate M be +Q and charge stored on plate N be –Q.
• The magnitude of charge on the capacitor will be Q.

$$C=\dfrac{Q}{\Delta V}$$

$$Q= C\Delta V$$

Note- Whatever be the initial charge, final charge remains the same.

#### Calculate the magnitude of charge on each plate of a parallel plate capacitor of capacitance $$C=7\,\mu F$$ when connected across battery with emf  $$V=7\,V.$$

A $$60\,\mu C$$

B $$29\,\mu C$$

C $$49\,\mu C$$

D $$59\,\mu C$$

×

Charge on capacitor is given as

$$Q=CV$$

Given  $$C=7\,\mu F,\,V=7\,V$$

$$Q = 7\times10^{-6}\times7$$

$$Q = 49\times10^{-6}$$

$$Q = 49\,\mu C$$

### Calculate the magnitude of charge on each plate of a parallel plate capacitor of capacitance $$C=7\,\mu F$$ when connected across battery with emf  $$V=7\,V.$$

A

$$60\,\mu C$$

.

B

$$29\,\mu C$$

C

$$49\,\mu C$$

D

$$59\,\mu C$$

Option C is Correct

# Electric field due to Parallel Plate Capacitor

• Consider two parallel metallic plates of equal area carrying a charge of equal magnitude and opposite nature are separated by a distance $$d$$.
• The surface charge density on each plate is $$\sigma$$

where $$\sigma=\dfrac{Q}{A}$$

• When plates are placed very closed to each other (compared to length and width) electric field is assumed to be uniform.

• Since distance between plates is small, it can be considered as single conductor.

By Gauss's law,

Flux through Gaussian surface

$$\phi _E=\oint\,E.dA=\dfrac{q_{in}}{\epsilon_0}$$

or,  $$\phi _E=2 EA$$   ....(1)

Also,   $$\phi _E=\dfrac{q_{in}}{\epsilon_0}$$ ....(2)

From (1) and (2)

$$2EA=\dfrac{\sigma A}{\epsilon_0}$$   $$[q_{in}=\sigma\times A]$$

or, $$E_+=\dfrac{\sigma}{2\epsilon_0}$$

$$E_+=\dfrac{Q}{2\epsilon_0A}$$

Similarly for negative charge plate

$$E_-=\dfrac{Q}{2\epsilon_0A}$$

Total field =$$E_++E_-$$

Total field $$=\dfrac{Q}{2\epsilon_0A}+\dfrac{Q}{2\epsilon_0A}=\dfrac{Q}{\epsilon_0A}$$

#### Calculate electric field between plates of parallel plate capacitor having area  $$A=1\,m^2$$ and charge on each plate $$Q=8.85\,nC.$$

A $$10^2\,N/C$$

B $$10^5\,N/C$$

C $$10^4\,N/C$$

D $$10^3\,N/C$$

×

Electric field due to parallel plate capacitor is given as

$$E=\dfrac{Q}{\epsilon_0A}$$

Given : $$Q=8.85\,nC\,\,,\,\,A=1\,m^2$$

$$E=\dfrac{8.85\times 10^{-9}}{8.85\times 10^{-12}\times1}$$

$$E=10^3\,N/C$$

### Calculate electric field between plates of parallel plate capacitor having area  $$A=1\,m^2$$ and charge on each plate $$Q=8.85\,nC.$$

A

$$10^2\,N/C$$

.

B

$$10^5\,N/C$$

C

$$10^4\,N/C$$

D

$$10^3\,N/C$$

Option D is Correct