## 1. Figure Drawn below Shows a Capacitor Made of Two Circular Plates Each of Radius

12 cm12 cm

, and Separated by

5.0 cm5.0 cm

. The capacitor is being Charged by an External Source (not Shown in the Figure). The Charging Current is Constant and Equal to

0.15 A0.15 A

## 1. Calculate the Capacitance and the Rate of Change of the Potential Difference Between the Plates.

### Ans: Radius of each circular plates,

r=12 cm=0.12 mr=12 cm=0.12 m

Distance between the given plates,

d=5 cm=0.05 md=5 cm=0.05 m

Charging current, I=0.15AI=0.15A

The permittivity of free space, ε0=8.85×10−12C2N−1m−2ε0=8.85×10−12C2N−1m−2

Capacitance between the two plates can be given as:

C=ε0AdC=ε0Ad

Where,

A=πr2A=πr2.

Hence,

C=ε0πr2dC=ε0πr2d

C=8.85×10−12×(0.12)20.05C=8.85×10−12×(0.12)20.05

C=8.0032×10−12FC=8.0032×10−12F

C=80.032pFC=80.032pF

Charge on each plate is given as,

q=CVq=CV

, where,

V = Potential difference across the plates

Differentiating both sides with respect to time (t), we get:

dqdt=CdVdtdqdt=CdVdt

As, dqdt=Idqdt=I, so, the rate of change of potential difference between the plates can be given as:

dVdt=IC=0.1580.032×10−12dVdt=IC=0.1580.032×10−12

∴dVdt=1.87×109V/s∴dVdt=1.87×109V/s

## 2. Obtain the Displacement Current Across the Plates.

### Ans: The displacement current across the plates would be the same as the conduction current. Hence, the displacement current, IdId would be

0.15 A0.15 A

.

## 3. Is Kirchhoff’s First Rule (junction Rule) Valid at Each Plate of the Capacitor? Explain.

### Ans: Yes, Kirchhoff’s first rule would be valid at each plate of the capacitor, provided that sum of conduction and displacement currents, i.e., I=Ic+IdI=Ic+Id (junction rule of Kirchhoff’s law).

### 2) A Parallel Plate Capacitor (figure) Made of Circular Plates Each of Radius

R=6.0 cmR=6.0 cm

has a capacitance

C=100 pFC=100 pF

. The Capacitor is Connected to a

230 V230 V

ac supply with a (angular) frequency of

300rads−1300rads−1

## 1. What is the RMS Value of the Conduction Current?

### Ans: Radius of each circular plate,

R=6.0 cm=0.06 mR=6.0 cm=0.06 m

Capacitance of a parallel plate capacitor,

C=100 pF=100×10−12FC=100 pF=100×10−12F

Supply voltage,

V=230 VV=230 V

Angular frequency, ω=300rads−1ω=300rads−1

Rms value of conduction current can be given as: I=VXCI=VXC.

Where, XCXC = capacitive currant =1ωC=1ωC

∴I=V×ωC∴I=V×ωC

I=230×300×100×10−12I=230×300×100×10−12

I=6.9×10−6I=6.9×10−6

I=6.9μAI=6.9μA

Therefore, the RMS value of conduction current will be 6.9μA6.9μA.

## 2. Is the Conduction Current Equal to the Displacement Current?

### Ans: Yes, the conduction current will be the same as the displacement current. i.e., the conduction current is equal to the displacement current.

## 3. Determine the Amplitude of B→B→ at a point

3.0 cm3.0 cm

from the Axis Between the Plates.

### Ans: Magnetic field is given as: B→=μ0r2πR2I0B→=μ0r2πR2I0

Where,

μ0μ0= Free space permeability =4π×10−7NA−2=4π×10−7NA−2

I0I0 = Maximum value of current = 2–√I2I

rr= Distance between the plates from the axis = 3.0cm=0.03cm3.0cm=0.03cm.

∴B→=4π×10−7×0.03×2√×6.9×10−62π×(0.06)2∴B→=4π×10−7×0.03×2×6.9×10−62π×(0.06)2

B→=1.63×10−11TB→=1.63×10−11T

Therefore, the magnetic field at that point will be 1.63×10−11T1.63×10−11T.

## 3. What Physical Quantity is the Same for X-Rays of Wavelength

10−10m10−10m

, red light of wavelength

6800A∘6800A∘

and radio waves of wavelength

500 m500 m

?

### Ans: The speed of light (

3×108m/s3×108m/s

) is independent of the wavelength in the vacuum. Hence, it is the same for all wavelengths in a vacuum.

## 4. A Plane Electromagnetic Wave Travels in Vacuum Along Z-Direction. What Can You Say About the Directions of Its Electric and Magnetic Field Vectors? If the Frequency of the Wave is

30MHz30MHz

, what is its wavelength?

### Ans: The electromagnetic wave is travelling along the z-direction, in a vacuum. The electric field (E) and the magnetic field (H) will lie in the x-y plane and they will be mutually perpendicular.

Frequency of the wave,

ν=30 Mhz=30×106s−1ν=30 MHz=30×106s−1

Speed of light in a vacuum,

c=3×108m/sc=3×108m/s

Wavelength, λλ of the wave can be given as: λ=cνλ=cν

∴λ=3×10830×106=10m∴λ=3×10830×106=10m

## 5. A Radio Can Tune in to Any Station in the

7.5 Mhz7.5 MHz

to

12 Mhz12 MHz

Band. What is the corresponding Wavelength Band?

### Ans: A radio can tune to minimum frequency,

ν1=7.5 Mhz=7.5×106Hzν1=7.5 MHz=7.5×106Hz

Maximum frequency,

ν2=12 Mhz=12×106Hzν2=12 MHz=12×106Hz

Speed of light,

c=3×108m/sc=3×108m/s

Wavelength for frequency, ν1ν1 can be calculated as:

λ1=cν1λ1=cν1

λ1=3×1087.5×106λ1=3×1087.5×106

λ1=40mλ1=40m

Wavelength for frequency, ν2ν2 can be calculated as:

λ2=cν2λ2=cν2

λ2=3×10812×106λ2=3×10812×106

λ2=25mλ2=25m

Therefore, the corresponding wavelength band would be between 40m40m to 25m25m.

## 6. A Charged Particle Oscillates About Its Mean Equilibrium Position With a Frequency of 109Hz109Hz. What is the Frequency of the Electromagnetic Waves Produced by the Oscillator?

### Ans: The frequency of an electromagnetic wave produced by the oscillator will be equal to the frequency of charged particle oscillating about its mean position i.e.,

109Hz109Hz

.

## 7. The Amplitude of the Magnetic Field Part of a Harmonic Electromagnetic Wave in Vacuum is B0 = 510nT. What is the Amplitude of the Electric Field Part of the Wave?

### Ans: Amplitude of magnetic field of an electromagnetic wave in a vacuum, is:

B0=510nT=510×10−9TB0=510nT=510×10−9T

Speed of light in a vacuum, c=3×108m/sc=3×108m/s

Amplitude of electric field of the electromagnetic wave can be given as:

E=cB=3×108×510×10−9=153N/CE=cB=3×108×510×10−9=153N/C.

## 8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120N/C and that its frequency is

ν=50.0 Mhzν=50.0 MHz

.

1. Determine, B0B0, ωω, kk, and λλ.

### Ans: Electric field amplitude, E0=120N/CE0=120N/C

Frequency of source,

ν=50.0MHz=50×106Hzν=50.0MHz=50×106Hz

Speed of light,

c=3×108m/sc=3×108m/s

Magnetic field strength can be given as:

B0=E0cB0=E0c

B0=1203×108B0=1203×108

B0=4×10−7TB0=4×10−7T

B0=400nTB0=400nT

Angular Frequency of Source Can be Given As:

ω=2πνω=2πν

ω=2π(50×106)ω=2π(50×106)

ω=3.14×108rad/sω=3.14×108rad/s

Propagation Constant Can be Given As:

k=ωck=ωc

k=3.14×1083×108k=3.14×1083×108

k=1.05rad/mk=1.05rad/m

Wavelength of Wave Can be Given As:

λ=cνλ=cν

λ=3×10850×106λ=3×10850×106

λ=6.0mλ=6.0m

## 2. Find Expressions for E and B.

### Ans: If the wave propagates in the positive x-direction. Then, the electric field vector would be in the positive y-direction and the magnetic field vector will lie in the positive z-direction. This is because all three vectors are mutually perpendicular.

Equation of electric field vector can be given as:

E→=E0sin(kx−ωt)j^E→=E0sin(kx−ωt)j^

E→=120sin[1.05x−3.14×108t]j^E→=120sin[1.05x−3.14×108t]j^

And, magnetic field vector can be given as:

B→=B0sin(kx−ωt)k^B→=B0sin(kx−ωt)k^

B→=(4×10−7)sin[1.05x−3.14×108t]k^B→=(4×10−7)sin[1.05x−3.14×108t]k^

## 9. The Terminology of Different Parts of the Electromagnetic Spectrum Is Given in the Text. Use the Formula E = hv (for Energy of a Quantum of Radiation: Photon) and Obtain the Photon Energy in Units of ev for Different Parts of the Electromagnetic Spectrum. in What Way are the Different Scales of Photon Energies That You Obtain Related to the Sources of Electromagnetic Radiation?

### Ans: The energy of photon can be given as:

E=hν=hcλE=hν=hcλ

Where,

hh = Planck’s constant =6.6×10−34Js=6.6×10−34Js

cc = speed of light =3×108m/s=3×108m/s

λλ = wavelength of radiation

∴E=6.6×10−34×3×108λJ∴E=6.6×10−34×3×108λJ

E=19.8×10−26λJE=19.8×10−26λJ

E=19.8×10−26λ×1.6×10−19eVE=19.8×10−26λ×1.6×10−19eV

E=12.375×10−7λeVE=12.375×10−7λeV

For Different Values of in an Electromagnetic Spectrum, Photon Energies

are Listed in Below Table:

λ(m)λ(m) | E(eV)E(eV) |

103103 | 12.375×10−1012.375×10−10 |

1 | 12.375×10−712.375×10−7 |

10−310−3 | 12.375×10−412.375×10−4 |

10−610−6 | 12.375×10−112.375×10−1 |

10−810−8 | 12.375×10112.375×101 |

10−1010−10 | 12.375×10312.375×103 |

10−1210−12 | 12.375×10512.375×105 |