- #1

mr_coffee

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Hello everyone, I did a lot 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.

(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?

Picutre here:http://img135.imageshack.us/img135/2295/hrw723266rd.gif [Broken]

Work:

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

Flux = EAcos

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

Flux = .007199

Pic here: http://img202.imageshack.us/img202/4117/hrw723275xe.gif [Broken]

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

(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:

http://img140.imageshack.us/img140/3248/hw8dk.jpg [Broken]

C/m

Work:

http://img140.imageshack.us/img140/9611/51zh.jpg [Broken]

(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: http://img140.imageshack.us/img140/3586/69wa.jpg [Broken]

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.

(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: http://img293.imageshack.us/img293/6738/996ym.jpg [Broken]

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

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.

Picture: http://img215.imageshack.us/img215/8562/102bi.jpg [Broken]

I said it has to be graph V.

Thanks everyone, any help would be great!

**#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:http://img135.imageshack.us/img135/2295/hrw723266rd.gif [Broken]

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: http://img202.imageshack.us/img202/4117/hrw723275xe.gif [Broken]

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:

http://img140.imageshack.us/img140/3248/hw8dk.jpg [Broken]

**#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:

http://img140.imageshack.us/img140/9611/51zh.jpg [Broken]

**#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: http://img140.imageshack.us/img140/3586/69wa.jpg [Broken]

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: http://img293.imageshack.us/img293/6738/996ym.jpg [Broken]

**#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: http://img215.imageshack.us/img215/8562/102bi.jpg [Broken]

I said it has to be graph V.

Thanks everyone, any help would be great!

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