# 16.2 Plane electromagnetic waves  (Page 2/5)

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A similar argument holds by substituting E for B and using Gauss’s law for magnetism instead of Gauss’s law for electric fields. This shows that the B field is also perpendicular to the direction of propagation of the wave. The electromagnetic wave is therefore a transverse wave, with its oscillating electric and magnetic fields perpendicular to its direction of propagation.

## The speed of propagation of electromagnetic waves

We can next apply Maxwell’s equations to the description given in connection with [link] in the previous section to obtain an equation for the E field from the changing B field, and for the B field from a changing E field. We then combine the two equations to show how the changing E and B fields propagate through space at a speed precisely equal to the speed of light.

First, we apply Faraday’s law over Side 3 of the Gaussian surface, using the path shown in [link] . Because ${E}_{x}\left(x,t\right)=0,$ we have

$\oint \stackrel{\to }{E}·d\stackrel{\to }{s}=\text{−}{E}_{y}\left(x,t\right)l+{E}_{y}\left(x+\text{Δ}x,t\right)l.$

Assuming $\text{Δ}x$ is small and approximating ${E}_{y}\left(x+\text{Δ}x,t\right)$ by

${E}_{y}\left(x+\text{Δ}x,t\right)={E}_{y}\left(x,t\right)+\frac{\partial {E}_{y}\left(x,t\right)}{\partial x}\text{Δ}x,$

we obtain

$\oint \stackrel{\to }{E}·d\stackrel{\to }{s}=\frac{\partial {E}_{y}\left(x,t\right)}{\partial x}\left(l\text{Δ}x\right).$

Because $\text{Δ}x$ is small, the magnetic flux through the face can be approximated by its value in the center of the area traversed, namely ${B}_{z}\left(x+\frac{\text{Δ}x}{2},t\right)$ . The flux of the B field through Face 3 is then the B field times the area,

${\oint }_{S}\stackrel{\to }{B}·\stackrel{\to }{n}dA={B}_{z}\left(x+\frac{\text{Δ}x}{2},t\right)\left(l\text{Δ}x\right).$

$\oint \stackrel{\to }{E}·d\stackrel{\to }{s}=-\frac{d}{dt}{\int }_{S}\stackrel{\to }{B}·\stackrel{\to }{n}dA.$

$\frac{\partial {E}_{y}\left(x,t\right)}{\partial x}\left(l\text{Δ}x\right)=-\frac{\partial }{\partial t}\left[{B}_{z}\left(x+\frac{\text{Δ}x}{2},t\right)\right]\left(l\text{Δ}x\right).$

Canceling $l\text{Δ}x$ and taking the limit as $\text{Δ}x=0$ , we are left with

$\frac{\partial {E}_{y}\left(x,t\right)}{\partial x}=-\frac{\partial {B}_{z}\left(x,t\right)}{\partial t}.$

We could have applied Faraday’s law instead to the top surface (numbered 2) in [link] , to obtain the resulting equation

$\frac{\partial {E}_{z}\left(x,t\right)}{\partial x}=-\frac{\partial {B}_{y}\left(x,t\right)}{\partial t}.$

This is the equation describing the spatially dependent E field produced by the time-dependent B field.

Next we apply the Ampère-Maxwell law (with $I=0$ ) over the same two faces (Surface 3 and then Surface 2) of the rectangular box of [link] . Applying [link] ,

$\oint \stackrel{\to }{B}·d\stackrel{\to }{s}={\mu }_{0}{\epsilon }_{0}\left(d\text{/}dt\right)\underset{S}{\int }\stackrel{\to }{E}·n\phantom{\rule{0.2em}{0ex}}da$

to Surface 3, and then to Surface 2, yields the two equations

$\frac{\partial {B}_{y}\left(x,t\right)}{\partial x}=\text{−}{\epsilon }_{0}{\mu }_{0}\frac{\partial {E}_{z}\left(x,t\right)}{\partial t},\phantom{\rule{0.2em}{0ex}}\text{and}$
$\frac{\partial {B}_{z}\left(x,t\right)}{\partial x}=\text{−}{\epsilon }_{0}{\mu }_{0}\frac{\partial {E}_{y}\left(x,t\right)}{\partial t}.$

These equations describe the spatially dependent B field produced by the time-dependent E field.

We next combine the equations showing the changing B field producing an E field with the equation showing the changing E field producing a B field. Taking the derivative of [link] with respect to x and using [link] gives

$\begin{array}{c}\hfill \frac{{\partial }^{2}{E}_{y}}{\partial {x}^{2}}=\frac{\partial }{\partial x}\left(\frac{\partial {E}_{y}}{\partial x}\right)=-\frac{\partial }{\partial x}\left(\frac{\partial {B}_{z}}{\partial t}\right)=-\frac{\partial }{\partial t}\left(\frac{\partial {B}_{z}}{\partial x}\right)=\frac{\partial }{\partial t}\left({\epsilon }_{0}{\mu }_{0}\frac{\partial {E}_{y}}{\partial t}\right)\hfill \\ \hfill \text{or}\hfill \end{array}$
$\frac{{\partial }^{2}{E}_{y}}{\partial {x}^{2}}={\epsilon }_{0}{\mu }_{0}\frac{{\partial }^{2}{E}_{y}}{\partial {t}^{2}}.$

This is the form taken by the general wave equation for our plane wave. Because the equations describe a wave traveling at some as-yet-unspecified speed c , we can assume the field components are each functions of x ct for the wave traveling in the + x -direction, that is,

define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
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pls. explain the lack of symmetry in the field of the parallel capacitor
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tell me about charging and discharging of capacitors
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equivalent capacitance is 9nf nd pd across each capacitor is 500v
santanu
four effect of heat on substances
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because you can't fit the boundary conditions.
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what is thermodynamics
the study of heat an other form of energy.
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heat is internal kinetic energy of a body but it doesnt mean heat is energy contained in a body because heat means transfer of energy due to difference in temperature...and in thermo-dynamics we study cause, effect, application, laws, hypothesis and so on about above mentioned phenomenon in detail.
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Total number of field lines crossing the surface area
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Basically flux in general is amount of anything...In Electricity and Magnetism it is the total no..of electric field lines or Magnetic field lines passing normally through the suface
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what is temperature change
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a bottle of soft drink was removed from refrigerator and after some time, it was observed that its temperature has increased by 15 degree Celsius, what is the temperature change in degree Fahrenheit and degree Celsius
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process whereby the degree of hotness of a body (or medium) changes
Salim
Q=mcΔT
Salim
where The letter "Q" is the heat transferred in an exchange in calories, "m" is the mass of the substance being heated in grams, "c" is its specific heat capacity and the static value, and "ΔT" is its change in temperature in degrees Celsius to reflect the change in temperature.
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what was the temperature of the soft drink when it was removed ?
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15 degree Celsius
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15 degree
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ok I think is just conversion
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15 degree Celsius to Fahrenheit
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0 degree Celsius = 32 Fahrenheit
Salim
15 degree Celsius = (15×1.8)+32 =59 Fahrenheit
Salim
I dont understand
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the question said you should convert 15 degree Celsius to Fahrenheit
Salim
To convert temperatures in degrees Celsius to Fahrenheit, multiply by 1.8 (or 9/5) and add 32.
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what is d final ans for Fahrenheit and Celsius
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it said what is temperature change in Fahrenheit and Celsius
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the 15 is already in Celsius
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So the final answer for Fahrenheit is 59
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what is d final ans for Fahrenheit and Celsius
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increase the capacitance.
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mechanical stiffness and small size
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so as to increase the capacitance of a capacitor
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