gauss's law (urgent)

2011-11-24 4:05 am

圖片參考：http://imgcld.yimg.com/8/n/HA01097219/o/701111230084713873404310.jpg

55. why the magnitude of E-field is (surface charge density)/(eo) but not (surface charge density)/2eo since a football field is a non-conductor?

56. Why Q' is the same as that of A (in part a, which I don't show it here)?

3. Why does a charge always move from high potential to low potential?

4. The magnitude of E-field between two thin, infinite conducting plate with surface charge density A is equal to A/e0, what about the an E-field between two thin, infinite conducting plate when one of them is charged and the other is not?

5. On an irregularly shaped conductor, why the surface charge density is greatest at locations where radius of curvature of the surface is smallest? Can some1 show the proof here or provide a website that contains the proof?

thanks

更新1:

6. A charge distribution that is spherically symmetric but not uniform radially produces an E-field of magnitude E=Kr^4, directed radially outward from the center of the sphere. r is the radial distance from that center, and K is a constant. What is the volume density p of the charge distribution?

更新2:

ans to 6: 6K(eo)r^3

更新3:

7. A solid nonconducting sphere of radius R = 5.6 cm has a nonuniform charge distribution of volume charge density p=(14.1 pC/m3)r/R, where r is radial distance from the sphere's center. a:

更新4:

(a) What is the sphere's total charge? What is the magnitude E of the electric field at (b) r=0, (c) r=R/2, (d) r=R? ans: a: 7.78fC, b: 0, c:5.58mN/C, d: 22.3mN/C

更新5:

1. So what's the differences between σ/e and σ/2e? 2. For 7a, why it is 4pi r^2 but not 4/3 pi r^3 as p is volume charge density. please send the ans to my email if there is not enough space here

回答 (1)

天同

2011-11-24 7:34 am

✔ 最佳答案

Due to limyed space here, I can only give you short answers.

55. The formula holds irrespective the football field is a conductor or not.

54. I think it is a misprint. Q' = p.(pi).r^2.h

3. Electric fields points from high to low potential. By definition of direction of electric field, +ve charge moves along field lines, thus it moves from high to low potential, similar to the situation that a mass moves from a highlevel to a low level along a slope.

4. The same equation applies. There is induced charge on the other plate even if it is originally not charged.

5. Potnetial V = k.Q/r
where k is a constant, Q is the charge and r is the radius of a spherical conductor
but Q = p.(4.pi.r^2), where p is the surface charge density
hence, V = k(4.pi).p.r
i.e. p = V/[k.(4.pi)r]
In a conductor, all surfaces are of equal potential. Hence, curvature with small radius r would have high surface density p.

6. Using Gauss Law, charge q enclosed in a sphere of radius r = (Kr^4).(4.pi.r^2)(e), where e is the permittivity of free space
hence, q = K.(4.pi.e).r^6
dq/dr = (24).(K.pi.e)r^5
But dq = (4.pi.r^2).dr.p where p is the volume charge density
i.e. dq/dr = (4.pi.r^2)p
Therefore, (24).(K.pi.e)r^5 = (4.pi.r^2)p
p = 6K.e.r^3

7.(a) total charge = integral[4.pi.r^2.p.dr
where p = 14.1r/R
hence total charge = 14.1.pi.R^3 = 14.1 x pi x (0.056)^3 pC/m^3 = 7.78x10^-15 C/m^3

(b) As in (a), E.(4.pi.r^2) = (14.1 x pi x r^4/R)/e
i.e. E = (14.1/4)r^2/(Re)
hence E = 0 N/C when r = 0 cm
you could substitute other values of r to find the electric field E

2011-11-24 21:42:00 補充：
1. So what's the differences between σ/e and σ/2e?
It depends on how you choose your Gaussian surface.

2

2011-11-24 21:44:11 補充：
7.(a) total charge = integral[4.pi.r^2.p.dr]
the term [4.pi.r^2.dr] is an elementary volume, [4.pi.r^2.dr]p is an elementary charge dq.
Integrate dq gives the total charge.

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