Hello everyone, I did alot of problems, but I want to make sure I did them correctly. If anyone could double check to see if I didn't make any stupid mistakes or if I totally screwed some up would be great. I'll post the questions and then Scan my work.(adsbygoogle = window.adsbygoogle || []).push({});

#1. A surface has the area vector A = (10i + 3j) m^2.

(a) What is the flux of an electric field through it if the field is E = 4 N/C i? The way I did this problem is, I found the magnitude of A sqrt(10^2+3^2) = 10.4 m^2 then I multiplied the A by 4 n/c to get 15.99 Nm^2/C. But now that I think of it, shouldn't I have just took 10i * 4, since the E field is only in the i direction and its the Dot product? So should it be 40 n/c?

(b) What is the flux of an electric field through it if the field is E = 4 N/C k, this would be 0, because the area vector doesn't even have a k component right?

#2.The square surface shown in Figure 23-26 measures 2.6 mm on each side. It is immersed in a uniform electric field with magnitude E = 1300 N/C. The field lines make an angle of = 35° with a normal to the surface, as shown. Take the normal to be directed "outward," as though the surface were one face of a box. Calculate the electric flux through the surface.

Picutre here:23-26

Work:

A = .0026^2 = 6.76E-6 m^2

Flux = EAcos

Flux = (1300N/C)(6.76E-6)(cos(35))

Flux = .007199

#3.At each point on the surface of the cube shown in Figure 23-27, the electric field is parallel to the z axis. The length of each edge of the cube is 4.0 m. On the top face of the cube E = -40 k N/C, and on the bottom face of the cube E = +29 k N/C. Determine the net charge contained within the cube.

Pic here: 23-27

Flux = (29E3 N/C)(4.0)^2;

Flux = 464000 k going up

Flux = (-40E3 N/C)(4.0)^2;

Flux = -640000 k going down

Net Flux = -640000 + 464000 = -176000

#4.A uniformly charged conducting sphere of 0.9 m diameter has a surface charge density of 7.8 µC/m2.

(a) Find the net charge on the sphere.

C

(b) What is the total electric flux leaving the surface of the sphere?

Nm2/C

Work:

Here

#5.An infinite line of charge produces a field of magnitude 4.9 104 N/C at a distance of 1.6 m. Calculate the linear charge density.

C/m

Work:

here

#6.Two long, charged, thin-walled, concentric cylinders have radii of 3.0 and 6.0 cm. The charge per unit length is 4.9 10- 6 C/m on the inner shell and -8.5 10-6 C/m on the outer shell.

(a) Find the magnitude and direction of the electric field at radial distance r = 4.8 cm from the common central axis. (Take radially outward to be positive.)

7.74E-6 N/C

(b) Find the magnitude and direction of the electric field at r = 8.5 cm, using the same sign convention.

7.74E-6 N/C

Work: Here

I first found the E field of the inner cylinder and outter cylinder. By using:

E = [tex]\gamma[/tex]/2PI*Eor;

Inner Cylinder:

E = 4.9E-6/(2PI(8.85E-12)(.03m) = 2.9E6 N/C;

Outter Cylinder:

E = -8.5E-6/(2PI(8.85E-12)(.06m) = -2.5E6 N/C;

You can see how I got the answers from the scanned work now.

#7.Charge of uniform density = 3.6 µC/m3 fills a nonconducting solid sphere of radius 4.5 cm.

(a) What is the magnitude of the electric field 3.5 cm from the center of the sphere?

N/C

(b) What is it at 9.5 cm from the center of the sphere?

N/C

Work: here

#8.The flux of the electric field (24 N/C) i + (30 N/C) j + (16 N/C) k through a 2.0 m2 portion of the yz plane is:

60 N m2/C

48 N m2/C

34 N m2/C

42 N m2/C

32 N m2/C

I drew a picture and i thought it would be 24 N/C * 2.0 = 48

#9.Consider Gauss's law: E dA = q/0. Which of the following is true?

If the charge inside consists of an electric dipole, then the integral is zero

If q = 0 then E = 0 everywhere on the Gaussian surface

If a charge is placed outside the surface, then it cannot affect E on the surface

E must be the electric field due to the enclosed charge

On the surface E is everywhere parallel to dA

I said On the surface E is everywhere parallel to dA.

#10.Which of the following graphs represents the magnitude of the electric field as a function of the distance from the center of a solid charged conducting sphere of radius R?

Picture: Here

I said it has to be graph V.

Thanks everyone, any help would be great!

**Physics Forums - The Fusion of Science and Community**

# Can someone double check to see if I did these problems correctly? E-Flux problems

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Can someone double check to see if I did these problems correctly? E-Flux problems

Loading...

**Physics Forums - The Fusion of Science and Community**