Maxwell related equations converted from MKSA units to Gaussian units

In summary, the conversation discusses the conversion of Maxwell's equations from MKSA units to Gaussian units and vice versa. The equations involve the electric field, charge density, permittivity of free space, vector and scalar fields, and derivatives. The conversation also mentions helpful resources for understanding and writing out the equations in the desired units.
  • #1
Ed Quanta
297
0
So here is an example of what I am trying to do.

We know that div of an E field=p/eo

where p=charge density and eo=the permittivity of free space. This equation is expressed in MKSA units. In order to convert this into Gaussian units, we must multiply E by 1/sqare root of 4*pi*eo, and multiply p by square root of 4*pi*eo.

Thus we are left with E= p*4*pi in Gaussian units

Now where A is a vector field and T is a potential scalar, I know that B=curl of A

and E=-gradient of T -1/c*dA/dt in the Gaussian unitss. I have to then convert this into MKSA. I am not sure what to do here because unlike the example which I dealt with above, I do not know how to convert both sides of the equation accordingly. Help anyone?
 
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  • #2
This might help.
http://electron6.phys.utk.edu/phys594/Tools/e&m/summary/maxwell/maxwell.html [Broken]
( http://electron6.phys.utk.edu/phys594/ [Broken] for the main page)

I once tried to come up with a scheme to write the equations in a way that easily showed the conversion.

For example, I wrote [itex]\nabla\cdot E= \left[ \frac{1}{4\pi\epsilon_0}\right]4\pi \rho [/itex]. However, I never had the time to check that the whole scheme applied to all of electromagnetism was consistent.
 
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1. What are the differences between MKSA and Gaussian units?

The MKSA (meter-kilogram-second-ampere) system is a modernized version of the metric system, while Gaussian units are based on the centimeter-gram-second system. The main difference is in the unit of electric charge, where MKSA uses the Coulomb and Gaussian units use the Statcoulomb. Additionally, MKSA units are defined in terms of physical constants, while Gaussian units are defined in terms of electromagnetic constants.

2. Why do we convert Maxwell equations from MKSA to Gaussian units?

There are a few reasons for converting Maxwell equations from MKSA to Gaussian units. One reason is that Gaussian units have simpler equations, making it easier to work with in certain theoretical calculations. Additionally, Gaussian units are commonly used in certain fields of physics, such as plasma physics and quantum electrodynamics.

3. How do you convert from MKSA to Gaussian units?

The conversion from MKSA to Gaussian units involves using conversion factors for each unit. For example, the conversion factor for the unit of electric current (ampere to abampere) is √(4π), and for the unit of electric charge (Coulomb to Statcoulomb) is 1/√(4π). These conversion factors are applied to each term in the Maxwell equations.

4. What are the implications of using Gaussian units for Maxwell equations?

The use of Gaussian units can change the numerical values of the Maxwell equations, but the physical meaning and relationships between the equations remain the same. However, care must be taken when comparing results obtained using different unit systems, as different conventions may lead to different numerical values.

5. Are there any practical applications for using Gaussian units in Maxwell equations?

Yes, there are several practical applications for using Gaussian units in Maxwell equations. As mentioned earlier, Gaussian units are commonly used in certain fields of physics, such as plasma physics and quantum electrodynamics. This is because the simpler equations make it easier to perform theoretical calculations in these areas. Additionally, Gaussian units are also used in engineering applications, such as in the design of electromagnetic devices and circuits.

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