Answer :
### Question Analysis
We need to calculate three things:
1. The maximum electric field strength [tex]\( E \)[/tex] produced.
2. The corresponding maximum magnetic field strength [tex]\( B \)[/tex] in the electromagnetic wave.
3. The wavelength [tex]\( \lambda \)[/tex] of the electromagnetic wave.
Given values are:
- Potential difference, [tex]\( V = 4.00 \: \text{mV} = 4.00 \times 10^{-3} \: \text{V} \)[/tex]
- Distance, [tex]\( d = 0.350 \: \text{m} \)[/tex]
- Frequency, [tex]\( f = 1.00 \: \text{Hz} \)[/tex]
- Speed of light in vacuum, [tex]\( c = 3.00 \times 10^8 \: \text{m/s} \)[/tex]
### Solutions
1. Maximum Electric Field Strength [tex]\( E \)[/tex]
[tex]\[ E = \frac{V}{d} \][/tex]
Substituting the given values:
[tex]\[ E = \frac{4.00 \times 10^{-3} \: \text{V}}{0.350 \: \text{m}} = 0.01142857142857143 \: \text{V/m} \][/tex]
Therefore,
[tex]\[ E \approx 0.0114 \: \text{V/m} \][/tex]
2. Maximum Magnetic Field Strength [tex]\( B \)[/tex]
The relationship between the electric field [tex]\( E \)[/tex] and the magnetic field [tex]\( B \)[/tex] in an electromagnetic wave is given by:
[tex]\[ E = c \times B \][/tex]
Solving for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{E}{c} \][/tex]
Substituting the value of [tex]\( E \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ B = \frac{0.01142857142857143 \: \text{V/m}}{3.00 \times 10^8 \: \text{m/s}} = 3.80952380952381 \times 10^{-11} \: \text{T} \][/tex]
Therefore,
[tex]\[ B = 3.80952380952381 \times 10^{-11} \: \text{T} \][/tex]
3. Wavelength [tex]\( \lambda \)[/tex]
The wavelength [tex]\( \lambda \)[/tex] of the electromagnetic wave is given by:
[tex]\[ \lambda = \frac{c}{f} \][/tex]
Substituting the value of [tex]\( c \)[/tex] and [tex]\( f \)[/tex]:
[tex]\[ \lambda = \frac{3.00 \times 10^8 \: \text{m/s}}{1.00 \: \text{Hz}} = 300000000.0 \: \text{m} \][/tex]
Therefore,
[tex]\[ \lambda = 3.00 \times 10^8 \: \text{m} \][/tex]
We need to calculate three things:
1. The maximum electric field strength [tex]\( E \)[/tex] produced.
2. The corresponding maximum magnetic field strength [tex]\( B \)[/tex] in the electromagnetic wave.
3. The wavelength [tex]\( \lambda \)[/tex] of the electromagnetic wave.
Given values are:
- Potential difference, [tex]\( V = 4.00 \: \text{mV} = 4.00 \times 10^{-3} \: \text{V} \)[/tex]
- Distance, [tex]\( d = 0.350 \: \text{m} \)[/tex]
- Frequency, [tex]\( f = 1.00 \: \text{Hz} \)[/tex]
- Speed of light in vacuum, [tex]\( c = 3.00 \times 10^8 \: \text{m/s} \)[/tex]
### Solutions
1. Maximum Electric Field Strength [tex]\( E \)[/tex]
[tex]\[ E = \frac{V}{d} \][/tex]
Substituting the given values:
[tex]\[ E = \frac{4.00 \times 10^{-3} \: \text{V}}{0.350 \: \text{m}} = 0.01142857142857143 \: \text{V/m} \][/tex]
Therefore,
[tex]\[ E \approx 0.0114 \: \text{V/m} \][/tex]
2. Maximum Magnetic Field Strength [tex]\( B \)[/tex]
The relationship between the electric field [tex]\( E \)[/tex] and the magnetic field [tex]\( B \)[/tex] in an electromagnetic wave is given by:
[tex]\[ E = c \times B \][/tex]
Solving for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{E}{c} \][/tex]
Substituting the value of [tex]\( E \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ B = \frac{0.01142857142857143 \: \text{V/m}}{3.00 \times 10^8 \: \text{m/s}} = 3.80952380952381 \times 10^{-11} \: \text{T} \][/tex]
Therefore,
[tex]\[ B = 3.80952380952381 \times 10^{-11} \: \text{T} \][/tex]
3. Wavelength [tex]\( \lambda \)[/tex]
The wavelength [tex]\( \lambda \)[/tex] of the electromagnetic wave is given by:
[tex]\[ \lambda = \frac{c}{f} \][/tex]
Substituting the value of [tex]\( c \)[/tex] and [tex]\( f \)[/tex]:
[tex]\[ \lambda = \frac{3.00 \times 10^8 \: \text{m/s}}{1.00 \: \text{Hz}} = 300000000.0 \: \text{m} \][/tex]
Therefore,
[tex]\[ \lambda = 3.00 \times 10^8 \: \text{m} \][/tex]