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Ctrostatics, Gauss’s Law, and Potential

  1. Jan 16, 2006 #1
    electrostatics, Gauss’s Law, and Potential

    Hello. I need some help with the following problems.
    4. A solid spherical container has a radius of 15 cm. The electric field 30 cm from the center of this sphere has a magnitude of 800 N/C. What is the surface charge density (sigma) on the sphere?
    I did:
    EA= q/ e0
    E(4)(pi)(r^2) = [sigma(4)(pi)(r^2)] / e0
    E= sigma / e0
    800(8.85 X 10^-12) = sigma
    sigma = 7.08 X 10^-9
    But the answer is 2.8 X 10^ -8 C/m^2
    6. A uniform line charge of linear charge density lambda= 5 nC/m extends from x=0 to x=10 m. The magnitude of the electric field at the point y=12 m on the perpendicular bisector of the finite line of charge is?
    I did:
    I made a right triangle where 10 m was the length of the horizontal side and 12 m was the length of the vertical side. Then I found the hypotenuse to be 15.62 m
    E= lambda / [(2pi)(e0)(r)]
    E= (5X 10 ^ -9) / [(2pi)(8.85 X10^-112)(15.62)
    But the answer is 2.88 N/C
    16. An infinitely long cylindrical shell of radius 6.0 cm carries a uniform surface charge density sigma = 12 nC/m^2. The electric field at r = 6.1 cm is approximately?
    I did:
    EA = q/e0
    E(2)(pi)(r)(l) = [sigma(2)(pi)(r)(l) ] / e0
    E= sigma / e0
    E= 1355.93
    But the answer is 1.3 kN/c
    I don’t know if my answer is wrong or correct because of rounding.
    Any help would be greatly appreciated.
    Last edited: Jan 16, 2006
  2. jcsd
  3. Jan 16, 2006 #2
    For first question
    [tex]\oint \vec{E}\cdot d\vec{A}=\frac{q}{\epsilon}[/tex]
    so [tex] q=\epsilon EA [/tex] where [tex] {A}=4 \pi {R}^2 [/tex] R= 0.3 m
    [tex] \sigma =\frac {q}{4\pi {r}^2} [/tex]
    where r= 0.15 m
    Last edited: Jan 16, 2006
  4. Jan 16, 2006 #3
    For second question
    [tex] E= \frac{2}{4\pi \epsilon_0} \int_0^5 \frac{\lambda dx}{x^2+y^2}\frac{y}{\sqrt{x^2+y^2}} [/tex]

    both the answer of q1 and q2 are correct
    Last edited: Jan 16, 2006
  5. Jan 16, 2006 #4
    How did you get that equation? And what do I plug in for x and R? And what do you mean by "both the answer of q1 and q2 are correct"?
    Last edited: Jan 16, 2006
  6. Jan 16, 2006 #5
    I mean that the answer of q1 is [tex]2.8 \times 10^ -8 C/m^2 [/tex] and the anwer of q2 is [tex] 2.88 N/C[/tex]
    The R in the equetion represented for y which is 12m. I get this equetion by [tex] \int \frac{1}{4 \pi \epsilon_0} \frac{dq}{r^2}[/tex] where r is the distance between dq and [tex] (5,12) [/tex] and thus [tex]r^2=x^2 + y^2[/tex].
    And because of symmetery, we can notice that the horizontal component will be cancled so we just have to integrate the vertical component. That's why [tex]\frac{y}{\sqrt{x^2+y^2}} [/tex] appeared.
    Last edited: Jan 17, 2006
  7. Jan 17, 2006 #6
    For the last one, You seem to use the same radius. Electric filed should be

    E = sigma r / e.R where r= 6cm, and R= 6.1 cm. e=permitivity of free space.

    You will get your answer in N/C. To change to kN/C devide your answer by 1000.

    E = 1334 N/C or 1.3 kN/C.

    Note that you made the same mistake in your first problem where

    E should have been = r^2 sigma / R^2 e0

    Learn how to use the gauss' low again because it is important to know the basics.
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