Capacitor - Types of Capacitor and Capacitance

What is a Capacitor?

The capacitor is a device in which electrical energy can be stored. It is an arrangement of two-conductor generally carrying charges of equal magnitudes and opposite sign and separated by an insulating medium. The non-conductive region can either be an electric insulator or vacuum such as glass, paper, air or semi-conductor called as a dielectric.

Capacitor vary in shape and size, they have many important applications in electronics.

Related Physics Concepts:

What are Capacitors Used For?

  • Storing electric potential energy such as batteries.
  • Filtering out unwanted frequency signals
  • Delaying voltage changes when coupled with resistors.
  • Used as a sensing device.
  • Used in the audio system of the vehicle.
  • Used to separate AC and DC.

Capacitors

One of the conductors has a positive charge +Q and it is at potential +V. whereas the other has an equal negative charge, -Q and is at potential –V.

Watch this Video for More Reference

Charge on Capacitor

Note: Charge on the capacitor is Q.

Total charge/ the net charge on the capacitor is –Q + Q = 0.

Circuit Symbols

Circuit Symbol Of Capacitor Circuit Symbol Of Capacitor

Capacitance

The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates i.e. QαVQ\alpha V or Q = CV

The constant of proportionality (C) is termed as the capacitance of the capacitor.

Dimensional Formula and Unit of Capacitance

  • Unit of Capacitance: Farad (F)

The capacitor value can vary from a fraction of pico-farad to more than a micro Farad. Voltage level can range from a couple to a substantial couple of hundred thousand volts.

  • Dimensional Formula: M-1L-2I2T4

Commonly Used Scales

  • μF\mu F = 10-6F
  • nF = 10-9F
  • pF = 10+2F

Factors Affecting the Capacitance

Capacitance depends on the following factor:

  1. Shape and size of the conductor
  2. Medium between them
  3. Presence of other conductors near it.

Calculation of Capacitance

We will try to calculate the capacitance of differently shaped capacitors, the steps are followed;

  1. Assume the charge on the conductors(Q)
  2. Calculate the electric field between the plates (E)
  3. Calculate potential difference from electric field(V)
  4. Apply the relation, C=QVC=\frac{Q}{V}

Types of Capacitor

  • Parallel Plate Capacitor
  • Spherical capacitor
  • Cylindrical capacitor

Parallel Plate Capacitor

The parallel plate capacitor consists of two metal plates of Area, A and is separated by a distance d. The plate on the top is given a charge +Q and that at the bottom is given the charge –Q. A potential difference of V is developed between the plates.

The separation is very small compared to the dimensions of the plate so that the effect of bending outward of electric field lines at the edges and the non-uniformity of surface charge density at the edges can be ignored.

Parallel Plate Capacitor

The charge density on each plate of parallel plate capacitor has a magnitude of σ

σ = Q/A

From Gauss law, E = Q/ε0A

Also, E = V/d

Now taking field due to the surface charges, outside of the capacitor,

E=σ2ε0σ2ε0=0E = \frac{\sigma}{2\varepsilon _0}-\frac{\sigma}{2\varepsilon _0}=0

Inside  E=σ2ε0+σ2ε0=σε0=qAε0      Inside\;E = \frac{\sigma}{2\varepsilon _0}+\frac{\sigma}{2\varepsilon _0}=\frac{\sigma}{\varepsilon _0}=\frac{q}{A\varepsilon _0}\;\;\;

vd=qAε0\frac{v}{d} = \frac{q}{A\varepsilon _0}

or,C=qv=Aε0dor,C = \frac{q}{v} =\frac{A\varepsilon _0}{d}

This results is valid for vacuum between the capacitor plates. For other medium, then capacitance will be C=kAε0dC = \frac{kA\varepsilon _0}{d}, where k is the dielectric constant of the medium,

ε0=Permittivity  of  free  space=8.85×1012C2/Nm2\varepsilon _0 = Permittivity\;of\;free\;space = 8.85\times 10^{-12}C^{2}/N_{-m^{2}}

If there is vacuum between the plates, k=1.

Spherical Capacitor

Let’s consider a spherical capacitor that consists of two concentric spherical shells. Suppose the radius of the inner sphere, Rin = a and radius of the outer sphere, Rout = b. The inner shell is given a positive charge +Q and the outer shell is given –Q.

The charge enclosed by the sphere is Q. Thus by Gauss’s law,

E4πr2=Qε0E4\pi r^{2} = \frac{Q}{\varepsilon _0}

Spherical Capacitor

The potential difference in Spherical Capacitor between the two conducting shells is:

Δ=VhVa=ahErdr=Q4πε0ahdrr2=Q4πε0(1a1b)=Q4πε0(baab)\Delta = V_h-V_a=-\int_{a}^{h}E_rdr = -\frac{Q}{4\pi\varepsilon _0}\int_{a}^{h}\frac{dr}{r^{2}} = -\frac{Q}{4\pi\varepsilon _0}\left ( \frac{1}{a}-\frac{1}{b} \right ) =-\frac{Q}{4\pi\varepsilon _0}\left ( \frac{b-a}{ab} \right )

which yields,

C=QΔV=4πε0(abba)C = \frac{Q}{\left | \Delta V \right |} = 4\pi\varepsilon _0\left (\frac{ab}{b-a} \right )

Cylindrical Capacitor

Consider a solid cylinder of radius, a surrounded by a cylindrical shell, b. The length of the cylinder is l and is much larger than a-b to avoid edge effects. The capacitor is charged so that the charge on inner cylinder is +Q and outer cylinder is –Q.

From gauss’s law,

E=Q2πε0rl=λ2πε0rE = \frac{Q}{2\pi \varepsilon_0 rl } = \frac{\lambda}{2\pi\varepsilon _0r}

Where λ = Q/l, linear charge density

Cylindrical Capacitor

The potential difference of Cylindrical Capacitor is given by,

ΔV=VbVa=ahErdr=λ2πε0ln(ba)\Delta V = V_b – V_a = -\int_{a}^{h}E_rdr = -\frac{\lambda}{2\pi\varepsilon _0}\ln \left ( \frac{b}{a} \right )

Where we have chosen the integration path to be along the direction of the electric field lines. As expected, the outer conductor with negative charge has a lower potential. That gives

C=QΔV=λLλln(b/a)/rπε0=2πε0Lln(b/a)C = \frac{Q}{\left | \Delta V \right |} = \frac{\lambda L}{\lambda \ln (b/a)/r\pi\varepsilon _0} = \frac{2\pi\varepsilon _0L}{\ln(b/a)}

Once again, we see that the capacitance C depends only on the geometrical L, a and b.

Problems on Capacitor and Capacitance

Problem 1: Find the capacitance of a conducting sphere of radius R.

Sol: Let charge Q is given to sphere. The field outside the sphere at distance r is E = =kQr2=\frac{kQ}{r^{2}}

dVdr=E-\frac{dV}{dr} = E

0vdV=REdr\int_{0}^{v}dV = -\int_{\infty }^{R}Edr

V=kQ[1r]RV = kQ\left [ -\frac{1}{r} \right ]_{\infty }^{R}

V=kQRV = \frac{kQ}{R}

C=QV=R1/4πε0=4πε0RC = \frac{Q}{V} = \frac{R}{1/4\pi\varepsilon _0} = 4\pi\varepsilon _0R

 

Problem 2: A parallel plate air capacitor is made using two plates 0.2m square, spaced 1cm apart. It is connected to a 50V battery.

  1. What is the capacitance?
  2. What is the charge on each plate?
  3. What is the electric field between two plates?
  4. If the battery is disconnected and then the plates are pulled apart to a separation of 2cm, what are the answers to the above parts?

Sol:

  • Ca=ε0Ad0=8.85×1012×0.2×0.20.01C_{a} = \frac{\varepsilon_{0}A}{d_{0}} = \frac{8.85\times10^{-12}\times0.2\times0.2}{0.01}
  • Q=C0V0=(3.54×105×50)μC=1.77×105μCQ_=C_0V_0=(3.54\times 10^{-5}\times50)\mu C = 1.77\times10^{-5}\mu C
  • E0=V0d0=500.01=5000V/mE_0 = \frac{V_0}{d_0} =\frac{50}{0.01} = 5000V/m
  • If the battery is disconnected, the charge on the capacitor plates remains constant while the potential difference between plates can change.

C=Aε02d=1.77×105μfC = \frac{A\varepsilon_{0}}{2d} = {1.77\times10^{-5}}\mu f

Q=Q0=1.77×103μFQ=Q_0= 1.77\times10^{-3}\mu F

V=QC=Q0Cb/2=2V0=100voltsV = \frac{Q}{C} = \frac{Q_{0}}{C_{b}/2}=2V_{0} = 100 volts

E=VC=2V02d0=E0=5000V/mE = \frac{V}{C} = \frac{2V_{0}}{2d_{0}} = E_{0} = 5000 V/m

 

Problem 3: A parallel plate conductor connected in the battery with a plate area of 3.0 cm2 and plate separation is of 3mm if the charge stored on the plate is 4.0pc. Calculate the voltage of the battery?

Sol:

Area A = 3.0 cm2 = 3.0 × 10-4 m2

Ca=ε0Ad0C_{a} = \frac{\varepsilon_{0}A}{d_{0}}

 

Ca=ε0Ad0=8.85×1012(3×104)3×103C_{a} = \frac{\varepsilon_{0}A}{d_{0}} = \frac{8.85\times 10^{-12}\left ( 3\times 10^{-4} \right )}{3\times 10^{-3}}

 

C= 8.85 × 10-13

 

C=QVC = \frac{Q}{V}

 

V=QCV = \frac{Q}{C}

 

V=4×10128.85××1013V = \frac{4\times 10^{-12}}{8.85\times\times 10^{-13}}

 

V = 4.52 V

 

Dielectrics and Capacitance

What are Dielectrics?

Dielectrics and Capacitance

It is an insulating material (non-conducting) which has no free electrons. But a microscopic displacement of charges is observed in the presence of an electric field. It is found that the capacitance increases as the space between the conducting plates are filled with dielectrics.

Polar and Non-polar Dielectrics

Each atom is made of a positively charged nucleus surrounded by electrons. If the centre of the negatively charged electrons does not coincide with the centre of the nucleus, then a permanent dipole (separation of charges over a distance) moment is formed. Such molecules are called polar molecules. If a polar dielectric is placed in an electric field, the individual dipoles experience a torque and try to align along the field.

In non-polar molecules, the centres of the positive and negative charge distributions coincide. There is no permanent dipole moment created. But in the presence of an electric field, the centres are slightly displaced. This is called induced dipole moments.

Polar and Non-polar Dielectrics

Polarization of a Dielectric Slab

It is the process of inducing charges on the dielectric and creating a dipole moment. Dipole moment appears in any volume of a dielectric. The polarization vector p\overrightarrow{p} is defined as the dipole moment per unit volume.

Polarization of a Dielectric slab polarization in a dielectric slab

Dielectric Constant

Let E0\overrightarrow{{{E}_{0}}} be the electric field due to external sources and Ep\overrightarrow{{{E}_{p}}} be the field due to polarization (induced). The resultant field is

E=E0+Ep\overrightarrow{E}=\overrightarrow{{{E}_{0}}}+\overrightarrow{{{E}_{p}}}.

The induced electric field is opposite in direction to the applied field. But the resultant field is in the direction of the applied field with reduced magnitude.

E=E0K\overrightarrow{E}=\frac{\overrightarrow{{{E}_{0}}}}{K} K is called the dielectric constant or relative permittivity of the dielectric. For vacuum, Ep\overrightarrow{{{E}_{p}}} = 0, K = 1. It is also denoted by ε

Effect of Dielectric in Capacitance

Dielectric Slabs in Series

A parallel plate capacitor contains two dielectric slabs of thickness d1, d2 and dielectric constant k1 and k2 respectively. The area of the capacitor plates and slabs is equal to A.

Dielectric Slabs in Series

Considering the capacitor as combination of two capacitors in series, the equivalent capacitance C is given by:

1C=1C1+1C2\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}

1C=d1k1ε0A+d2k2ε0A\frac{1}{C}=\frac{{{d}_{1}}}{{{k}_{1}}\varepsilon 0A}+\frac{{{d}_{2}}}{{{k}_{2}}{{\varepsilon }_{0}}A}

C=ε0Ad1k1+d2k2C=\frac{{{\varepsilon }_{0}}A}{\frac{{{d}_{1}}}{{{k}_{1}}}+\frac{{{d}_{2}}}{{{k}_{2}}}}

Dielectric Slabs in Parallel

Dielectric slabs in parallel

Consider a capacitor with two dielectric slabs of same thickness d placed inside it as shown. The slabs have dielectric constants k1 and k2 and areas A1 and A2 respectively. Treating the combination as two capacitors in parallel,

C = C1 + C2

C = C=k1ε0A1d+k2ε0A2dC=ε0d[k1A1+k2A2]C=\frac{{{k}_{1}}{{\varepsilon }_{0}}{{A}_{1}}}{d}+\frac{{{k}_{2}}{{\varepsilon }_{0}}{{A}_{2}}}{d}\,\,\,\,\Rightarrow \,\,\,C=\frac{{{\varepsilon }_{0}}}{d}[{{k}_{1}}{{A}_{1}}+{{k}_{2}}{{A}_{2}}]

Dielectric and Vacuum

If there exits a dielectric slab of thickness t inside a capacitor whose plates are separated by distance d, the equivalent capacitance is given as:

C=ε0Atk+dt1(k=1forvacuum)C=\frac{{{\varepsilon }_{0}}A}{\frac{t}{k}+\frac{d-t}{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(k=1\,for\,vacuum)

C=ε0Atk+dtC=\frac{{{\varepsilon }_{0}}A}{\frac{t}{k}+d-t}

The equivalent capacitance is not affected by changing the distance of slab from the parallel plates. If the slab is of metal, the equivalent capacitance is: C=ε0AdtC=\frac{{{\varepsilon }_{0}}A}{d-t} (for a metal, k =)

Problems on Capacitance and Dielectrics

Problem 1: Three capacitors of 10μF each are connected as shown in the figure. Two of them are now filled with dielectric with K = 2, K = 2.5 as shown. Find the equivalent capacitance.

After insertion of dielectrics,

C1 = 10μF; C2 = KC0 = 2 x 10 = 20μF; C3 = KC0 = 2.5 x 10 = 25 μF

Ceff=10×2010+20+25=3123μF\,{{C}_{eff}}=\frac{10\times 20}{10+20}+25=31\frac{2}{3}\mu F

 

Problem 2: Find the equivalent capacitance of the system shown (assume square plates).

Taking K1 = 2 to be series in K2 = 3

1cleft=1(2)ε0{(L)(L3)}(d3)+1(3)ε0{(L)(L3)}(2d3)Cleft=6ε0L27d\Rightarrow \,\,\frac{1}{{{c}_{left}}}=\frac{1}{\frac{(2){{\varepsilon }_{0}}\left\{ (L)\left( \frac{L}{3} \right) \right\}}{\left( \frac{d}{3} \right)}}+\frac{1}{\frac{(3){{\varepsilon }_{0}}\left\{ (L)\left( \frac{L}{3} \right) \right\}}{\left( \frac{2d}{3} \right)}}\Rightarrow \,\,\,\,{{C}_{left}}=\frac{6{{\varepsilon }_{0}}{{L}^{2}}}{7d} Now

Cright=(4)ε0{(L)(2L3)}d=8ε0L23d{{C}_{right}}=\frac{(4){{\varepsilon }_{0}}\left\{ (L)\left( \frac{2L}{3} \right) \right\}}{d}=\frac{8{{\varepsilon }_{0}}{{L}^{2}}}{3d}

Now Cleft and Cright are in parallel

Ceq=Cleft+Cright=6ε0L27d+8ε0L23d=74ε0L221d\Rightarrow \,\,\,{{C}_{eq}}={{C}_{left}}+{{C}_{right}}=\frac{6{{\varepsilon }_{0}}{{L}^{2}}}{7d}+\frac{8{{\varepsilon }_{0}}{{L}^{2}}}{3d}=\frac{74{{\varepsilon }_{0}}{{L}^{2}}}{21d}

 

Problem 3: Calculate the effective capacitance connected in series and parallel? And the capacitors connected to 40 V battery. Calculate the voltage across the capacitors for each connection type.

Given

C1 = 12 F

C2 = 6 F

Solu:

When capacitors are connected in series

1C=1C1+1C2\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

 

1C=112+16\frac{1}{C} = \frac{1}{12} + \frac{1}{6}

 

1C=0.25\frac{1}{C} = 0.25

 

C = 4 F

C=QVC = \frac{Q}{V}

Q = CV

Q = 4 × 40

Q = 160 C

For the 12 F capacitor:

12=160V12 = \frac{160}{V}

V = 13.33V

For the 6F capacitor:

6=160V6 = \frac{160}{V}

V = 26.66V

When capacitors are connected in parallel

C = C1 + C2

C = 12 + 6

C = 18 F

The volatege is same as 40V across the each capacitors.

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