# Ctrostatics, Gauss’s Law, and Potential

1. Jan 16, 2006

### Shay10825

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)
E=7.49
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.
Thanks

Last edited: Jan 16, 2006
2. Jan 16, 2006

### Psi-String

For first question
$$\oint \vec{E}\cdot d\vec{A}=\frac{q}{\epsilon}$$
so $$q=\epsilon EA$$ where $${A}=4 \pi {R}^2$$ R= 0.3 m
thus
$$\sigma =\frac {q}{4\pi {r}^2}$$
where r= 0.15 m

Last edited: Jan 16, 2006
3. Jan 16, 2006

### Psi-String

For second question
$$E= \frac{2}{4\pi \epsilon_0} \int_0^5 \frac{\lambda dx}{x^2+y^2}\frac{y}{\sqrt{x^2+y^2}}$$

both the answer of q1 and q2 are correct

Last edited: Jan 16, 2006
4. Jan 16, 2006

### Shay10825

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
5. Jan 16, 2006

### Psi-String

I mean that the answer of q1 is $$2.8 \times 10^ -8 C/m^2$$ and the anwer of q2 is $$2.88 N/C$$
The R in the equetion represented for y which is 12m. I get this equetion by $$\int \frac{1}{4 \pi \epsilon_0} \frac{dq}{r^2}$$ where r is the distance between dq and $$(5,12)$$ and thus $$r^2=x^2 + y^2$$.
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 $$\frac{y}{\sqrt{x^2+y^2}}$$ appeared.

Last edited: Jan 17, 2006
6. Jan 17, 2006

### Gamma

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.