Displacement current inside spherical capacitor

In summary, the conversation discusses a spherical capacitor with inner and outer conductor radii, filled with a perfect homogeneous dielectric and connected to a low-frequency time-harmonic voltage. The displacement current density vector at an arbitrary point in the dielectric is found by computing the capacitance, using the current equation, and dividing by the surface area at that point. Gauss's law and the charge-to-voltage ratio are used to calculate the capacitance.
  • #1
Mr.Tibbs
24
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You have a spherical capacitor with inner conductor radius a and outer conductor with radius b. The capacitor is filled with a perfect homogeneous dielectric of permittivity ε and is connected to a low-frequency time-harmonic voltage v(t)=V[itex]_{0}[/itex]cos(ωt). Find the displacement current density vector at an arbitrary point in the dielectric.

Equations:

C=[itex]\frac{εS}{r}[/itex] ; S-area of the plates; r-distance

i(t)=C[itex]\frac{dv}{dt}[/itex]


Approach:

for starters I subbed the capacitance equation into the current equation and achieved this result:

i(t)=[itex]\frac{εS}{r}[/itex]*[itex]\frac{dv}{dt}[/itex]

taking the first derivative of the voltage and subbing it into the equation gives me:

i(t)=[itex]\frac{εS}{r}[/itex]*-V[itex]_{0}[/itex]ωsin(ωt)

Now I divide both sides of the equation by S in order to get the current density. I then integrate this equation with respect to r from the inner radius to the outer radius:

J[itex]_{d}[/itex](t)=-[itex]\frac{εV_{0}sin(ωt)}{ln(b/a)}[/itex]

More or less I don't know if I computed this correctly so any help would be appreciated.
 
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  • #2
Mr.Tibbs said:
You have a spherical capacitor with inner conductor radius a and outer conductor with radius b. The capacitor is filled with a perfect homogeneous dielectric of permittivity ε and is connected to a low-frequency time-harmonic voltage v(t)=V[itex]_{0}[/itex]cos(ωt). Find the displacement current density vector at an arbitrary point in the dielectric.

Equations:

C=[itex]\frac{εS}{r}[/itex] ; S-area of the plates; r-distance
Irrelevant. You do not have a parallel-plate capacitor.


i(t)=C[itex]\frac{dv}{dt}[/itex]
Approach:

.

1. Compute C for your capacitor. Don't forget ε.
2. Use i = CdV/dt to get the total current.
3. Find the current density at any point r, a < r < b, by dividing the total current by the area of the surface at r.

For (1), use Gauss's law and C = q/V. V = ∫abE(r)dr.
 

Related to Displacement current inside spherical capacitor

1. What is displacement current?

Displacement current is an electric current that is not caused by the flow of charge carriers, but instead by a changing electric field. It was first described by James Clerk Maxwell in his famous equations.

2. What is a spherical capacitor?

A spherical capacitor is a type of capacitor that consists of two concentric spheres separated by a dielectric material. It is a common component in electronic circuits and is used to store and release electrical energy.

3. How is displacement current calculated inside a spherical capacitor?

The displacement current inside a spherical capacitor can be calculated using the formula Id = ε0A(dE/dt), where Id is the displacement current, ε0 is the permittivity of free space, A is the area of the spherical capacitor, and dE/dt is the rate of change of the electric field.

4. What factors affect the displacement current inside a spherical capacitor?

The displacement current inside a spherical capacitor is affected by the permittivity of the dielectric material between the two spheres, the area of the capacitor, and the rate of change of the electric field.

5. How does displacement current affect the capacitance of a spherical capacitor?

Displacement current plays a crucial role in the overall capacitance of a spherical capacitor. It can increase or decrease the total capacitance depending on the direction and magnitude of the current. In general, displacement current increases the capacitance of a spherical capacitor.

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