familiari nf3795 Gauss Potential HW yeazell 58395 1 This print out should have 21 questions Multiple choice questions may continue on the next column or page nd all choices before answering 001 10 0 points A cubic box of side a oriented as shown con tains an unknown charge The vertically di rected electric eld has a uniform magnitude E at the top surface and 4 E at the bottom surface a E 4 E How much charge Q is inside the box 1 Qencl 0 2 Qencl 3 0 E a2 correct 3 insu cient information 4 Qencl 2 E 0 a2 5 Qencl 2 0 E a2 6 Qencl 6 0 E a2 7 Qencl 0 E a2 8 Qencl 9 Qencl 0 E a2 E 0 a2 1 2 E 0 a2 10 Qencl 3 Explanation out of the surface S and negative for lines go ing in No ux passes through the vertical sides The top receives top E a2 inward is negative and the bottom bottom 4 E a2 so the total electric ux is E E a2 4 E a2 3 E a2 Using Gauss s Law the charge inside the box is Qencl 0 E 3 0 E a2 002 part 1 of 6 10 0 points A positive test charge is placed at the center of a spherical Gaussian surface What happens to the net ux through the Gaussian surface when the surface is replaced by a cube of the same volume whose center is at the same point 1 No change correct 2 The net ux increases 3 The net ux decreases but is nonvanish ing 4 The net ux is zero Explanation By Gauss Law S IS E d A Q 0 As long as the charge remains the same and it is enclosed by any Gaussian surface the net ux will remain the same 003 part 2 of 6 10 0 points What happens to the net ux through the Gaussian surface when the sphere is replaced by a cube of one third the volume centered at the same point Electric ux through a surface S is by con vention positive for electric eld lines going 1 The net ux is zero 2 The net ux increases familiari nf3795 Gauss Potential HW yeazell 58395 2 the Gaussian surface when a second positive charge is placed near but outside the original sphere 3 No change correct 4 The net ux decreases but is nonvanish ing Explanation 004 part 3 of 6 10 0 points What happens to the net ux through the Gaussian surface when the charge is moved o center in the original sphere but remains within the sphere ing 1 The net ux is zero 2 The net ux increases 3 No change correct 4 The net ux decreases but is nonvanish 1 The net ux is zero 2 The net ux increases 3 No change correct 4 The net ux decreases but is nonvanish ing Explanation 005 part 4 of 6 10 0 points What happens to the net ux through the Gaussian surface when the charge is moved just outside the original sphere 1 No change 2 The net ux increases ing Explanation If the charge is moved outside the Gaussian surface then the amount of enclosed charge is zero and the net ux is zero One can imagine the eld lines going out from the charge Any line entering the closed Gaussian surface will also leave so the net ux is zero 006 part 5 of 6 10 0 points The positive test charge is again placed at the center of a spherical Gaussian surface What happens to the net ux through Explanation From Part 4 it is clear that an exterior charge does not contribute to the ux so the ux remains the same 007 part 6 of 6 10 0 points What happens to the net ux through the Gaussian surface when a second positive charge is placed inside the original spherical Gaussian surface 1 The net ux decreases but is nonvanish ing 2 The net ux is zero 3 The net ux increases correct Explanation ux must increase Since the net enclosed charge increases the 008 10 0 points A uniformly charged conducting plate with area A has a total charge Q which is positive Consider a cross sectional view of the plane and the electric eld lines due to the charge on the plane 3 The net ux is zero correct 4 The net ux decreases but is nonvanish 4 No change familiari nf3795 Gauss Potential HW yeazell 58395 3 E E cross section of the surface is S then Gauss Law states that Q P Find the magnitude of the eld at point P which is a distance a from the plate Assume that a is very small when compared to the dimensions of the plate such that edge e ects can be ignored Q 4 0 A 0 Q A Q 0 A 2 0 Q A 0 Q a2 correct Q 4 0 a2 Q 2 0 A 4 0 a2 Q 4 0 a Q 1 2 3 4 5 6 7 8 9 E E E E E k k k k k k k k k k E k k k k k k k k E E E E Q 4 0 a 10 k k Explanation Consider the Gaussian surface Q E E S Due to the symmetry of the problem there is an electric ux only through the right and left surfaces and these two are equal If the TOTAL 2 E S 1 0 Q A Q 2 0 A S E 009 part 1 of 2 10 0 points A long coaxial cable consists of an inner cylin drical conductor with radius R1 and an outer cylindrical conductor shell with inner radius R2 and outer radius R3 as shown The ca ble extends out perpendicular to the plane shown The charge on the inner conductor per unit length along the cable is and the corresponding charge on the outer conductor per unit length is same magnitude but opposite in sign and 0 R2 R1 b R3 Find the magnitude of the electric eld at the point a distance r1 from the axis of the inner conductor where R1 r1 R2 1 E correct 2 2 0 r1 R1 4 0 r1 2 3 0 r1 2 0 R1 2 R1 4 0 r1 R1 3 0 r1 2 0 r1 2 2 2 E 3 E 4 E 5 E 6 E 7 E familiari nf3795 Gauss Potential HW yeazell 58395 4 Four charges are placed at the corners of a q q3 q4 square of side a with q1 q2 q where q is positive Initially there is no charge at the center of the square q q3 q q2 Pick a cylindrical Gaussian surface with the radius r1 and apply Gauss s law q E 2 r1 0 E 2 0 r1 010 part 2 of 2 10 0 points Find the magnitude of the electric eld at the point a distance r2 from the axis of the …