Problem with electric flux density

In summary, the conversation is about a problem involving electric charge and electric field. The solution manual uses the formula D(r)=Q/[4*PI*r^2] in parts (a) and (b), but there is confusion about the correct formula to use in part (c). The answer provided seems to be correct, but there is a discussion about the Gaussian surface area and how it affects the electric field. The conversation ends with a question about the relationship between the surface areas at r = 20 mm and r = 10 mm.
  • #1
bibo_dvd
37
0
hello guys :)
i need your help with this problem

i can understand that in (a) we will use the integral of row(v) * dv to get the charge
and in (b) we will use the formule D(r)=Q/[4*PI*r^2]

but in (C) should it be zero or what ??
and if it should be zero why in the solution manual it uses the formula of D(r)=Q[4*PI*(r^2)] again ?!

i don't know what is the right answer for this ..so i need your help ..Thanks guys :)

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  • #2
No, the answer you have -- ##32.5 \frac{nC}{m^{2}}## -- seems right. The only change in part c is that you're looking farther away from the charge distribution, so there's still going to be electric field there and so the flux density will be nonzero, but since you're farther away from the charge, the density is going to be less.
 
  • #3
The Gaussian surface area A at r = 20 mm is what multiple of the surface at r = 10 mm? Use that to determine D(20mm)*A(20mm) = D(10mm)*(A(10mm).
 

Related to Problem with electric flux density

1. What is electric flux density?

Electric flux density is a measure of the strength of an electric field in a given region. It is a vector quantity that describes the amount of electric flux passing through a unit area perpendicular to the direction of the electric field.

2. How is electric flux density calculated?

Electric flux density is calculated by dividing the total electric flux passing through a surface by the area of that surface. It is also equal to the product of the electric field strength and the permittivity of the medium in which the field exists.

3. What is the unit of electric flux density?

The unit of electric flux density is coulombs per square meter (C/m2). It can also be expressed in terms of volts per meter (V/m) or newtons per coulomb (N/C).

4. Why is there a problem with electric flux density?

The problem with electric flux density is that it can be difficult to measure accurately, especially in situations where the electric field is not uniform. This can lead to errors in calculations and predictions of electric behavior.

5. How is electric flux density related to electric potential?

Electric flux density and electric potential are related through the electric field strength. Electric field strength is equal to the negative gradient of electric potential, so as electric flux density increases, electric potential decreases and vice versa.

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