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Electric field inside the cavity of a charged sphere Application


Physics

Superposition principle states that if a single excitation is broken down into few constitutive components, total response is the sum of the responses to individual components. The use of the principle can be illustrated on the following electrostatic example.
Sphere of radiusspace a equals 1 m with an empty, spherical cavity of a radius b equals 0.25 m , has a positive volume charge density rho space equals space plus 10 to the power of negative 6 end exponent space C over m cubed. spaceThe center of the cavity is at the distance d equals 0.5 m spacefrom the center of the charged sphere (Figure 1).

Positively charged sphere with an off-centered cavity

Figure 1 - Positively charged sphere with an off-centered cavity

According to the superposition principle, total field inside the cavity can be found by adding up individual fields of:

  • A positively charged ( rho equals plus 10 to the power of negative 6 end exponent space C over m cubed space), thoroughly filled sphere with a radius a equals 1 m.
  • A negatively charged ( rho equals negative 10 to the power of negative 6 end exponent C over m cubed) sphere whose size and position match the cavity (Fig. 2).


Representation of an empty volume by a superposition of two opposite charge density domains

Figure 2 - Representation of an empty volume by a superposition of two opposite charge density domains

Field of any isolated, uniformly charged sphere in its interior at a distance r, can be calculated from Gauss’ Law:

 E 4 pi r squared equals 1 over epsilon subscript 0 space fraction numerator 4 over denominator 3 space end fraction pi r cubed space rho

Which yields for a positive sphere:   

 stack E subscript plus space space with rightwards arrow on top equals fraction numerator rho over denominator 3 epsilon subscript 0 space end fraction space stack r subscript 1 with rightwards arrow on top space


And for a negative sphere: 

 E with rightwards arrow on top subscript minus equals negative fraction numerator rho over denominator 3 epsilon subscript 0 end fraction stack r subscript 2 with rightwards arrow on top

Where vectorsr with rightwards arrow on top subscript 1  and r with rightwards arrow on top subscript 2  are as defined in Figure 3.

Relationship between the individual Electric field directions and the vector d ? representing the cavity offset


Figure 3 - Relationship between the individual Electric field directions and the vector  representing the cavity offset

Therefore, the total electric field in the cavity can be computed as:

  E with rightwards arrow on top equals E with rightwards arrow on top subscript plus space plus stack E subscript minus with rightwards arrow on top equals fraction numerator rho over denominator 3 epsilon subscript 0 end fraction r with rightwards arrow on top subscript 1 minus r with rightwards arrow on top subscript 2


E with rightwards arrow on top equals fraction numerator rho over denominator 3 epsilon subscript 0 end fraction d with rightwards arrow on top

From the last equation, it can be concluded that the electric field in the cavity is constant with a direction  stack d space with rightwards arrow on topand that its magnitude (for d equals 0.5 m and rho equals space 10 to the power of negative 6 end exponent C divided by m cubed) isspace E equals 18.832 fraction numerator k V over denominator m end fraction.  The field magnitude depends only on the value of the charge density and the distance by which the center of the cavity is offset from the center of the sphere.

Model

Creating an Air domain

In EMS, electromagnetic analysis requires modeling of the surrounding air regions, because very often, significant part of the electromagnetic field extends outside the parts of the simulated system. Once Solidworks part representing the air domain has been imported in the assembly, all the parts should be subtracted from it. To do so:          

  1. Select the Air part in the Solidworks feature manager
  2. Click Edit component1E in the Solidworks Assembly tab
  3. In Solidworks menu click Insert/Molds/Cavity
  4. In the cavity feature manager, select Charged sphere and Cavity as the Design Components
  5. Click OK 2E.

3D model of sphere with a spherical cavity together with surrounding air domain
Figure 4 - 
3D model of sphere with a spherical cavity together with surrounding air domain

The simulation is performed as the EMS Electrostatic studyE  . Air is used as a material for all parts.
(To see how to assign materials, see the “Computing capacitance of a multi-material capacitor” example).

Boundary conditions            

To simulate the electric field, a Charge density boundary condition should be assigned to the large sphere, and a Fixed voltage boundary condition should be assigned to the face of the Air region.

To assign a charge density to the Charged sphere

  1. In the EMS manger tree, Right-click on the Load/Restraint 3E, select Charge density 4E , then choose Volume.
  2. Click inside the Bodies Selection5E box and then select the Charged sphere.
  3. In the Charge Density tab, type 1e-006.
  4. Click OK2E .

To see how to assign 0 Volt to the face of the Air region, see “Force in a capacitor” example.

Results

To display the variation of the electric field along the axis that connects the center of the Charged sphere and the center of the cavity:

  1. In the EMS manger tree, Under Results 7E , right click on the Electric Field folder 8Eand select 2D Plot then choose Linear.         
  2. The 2D Electric Field Property Manager Page appears.
  3.  In the Select points tab, select the start and the end points.
  4. In the Number of Points tab, type 1000.
  5. Click OK2E

In the obtained curve (Figure 5), it’s clear that the electric field in the cavity is constant, and its value is E equals 18.797 space K space v o l t s space, which closely matches the theoretical result. The field in Figure 5 steadily increases with the radius until it meets the cavity r space equals space 250 m m and then remains unchanged through the cavity (till r space equals space 750 m m ). The field peaks at the surface of the sphere ( r space equals space 1000 m m) and then it drops with the square of radius.

Electric field vs radius

Figure 5 - Electric field vs radius
 

Reference

[1] http://jkwiens.com/2007/10/24/answer-electric-field-of-a-nonconducting-sphere-with-a-spherical-cavity/