Double-Checking Electric Field Problems

In summary, the group discussed various problems involving electric fields and flux. They also discussed Gauss's law and the relationship between electric field and distance from a charged conducting sphere. In problem #1, there was confusion about whether to use the dot product or the magnitude of the area vector. The correct answer was 40 N/C for the first part and 0 for the second part. In problem #6, there was confusion about the units of charge and electric field, and the correct answer was found by calculating the total charge in the inner cylinder and adding it to the total charge in the outer cylinder. In problem #8, the group discussed how to find the flux of an electric field through a given area. Finally, in problem #10
  • #1
mr_coffee
1,629
1
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.

#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
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

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

#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

#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
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


#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

I said it has to be graph V.

Thanks everyone, any help would be great! :smile:
 
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  • #2
1a: your second thought is correct: 4i.10i= 40

1b: Yes, the flux is 0. Notice that if you use your first method in 1 a, you would get the wrong answer!
 
  • #3
Thanks! I knew somthing wasn't right!
 
  • #4
Can someone give me some advice on #6 please, the professor told me to find the total charge in the inner cylinder, and that will be the answer to A, and then add the total charge in the inner to the outer and that will give me answer B, but my answers make no sense, how can charge be E? its not even the right units.
 

FAQ: Double-Checking Electric Field Problems

What is the purpose of double-checking electric field problems?

Double-checking electric field problems is important because it ensures the accuracy of calculations and eliminates errors. This is crucial in scientific research and experimentation as even small mistakes can lead to incorrect conclusions.

What are some common errors that can occur in electric field problems?

Some common errors in electric field problems include mathematical mistakes, incorrect units, and overlooking important data or variables. These errors can significantly affect the results and conclusions of the problem.

How can I double-check my electric field calculations?

To double-check your electric field calculations, you can use multiple methods such as solving the problem using a different approach, using online calculators or software, or verifying your results with a colleague or mentor.

What are some tips for avoiding errors in electric field problems?

To avoid errors in electric field problems, it is important to carefully read and understand the problem, use consistent units, and keep track of all the variables and equations used. It is also helpful to take breaks and review your work periodically to catch any mistakes.

Why is it important to double-check electric field problems in scientific research?

In scientific research, double-checking electric field problems is crucial to ensure the validity and accuracy of the results. This helps to avoid any flawed conclusions or faulty experiments, leading to more reliable and trustworthy scientific findings.

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