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. #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!