008 10.0 points Consider a conducting sphere with radius R and charge +Q, and a conducting spherical shell with inner radius 2 R, outer radius 3 R and net charge of +Q .+Q +Q Select the E(r) var (electric field vs radial distance) diagram which describes this situa- tion. The unit of measure on the vertical axis is a multiple of * Q 18 R2 . E 18 16 14 12 10 + 8+ 1. 6 + 4 2 R 2R 3R -2 E 18 . 16 14 12 10 8 2. 6+ 4 2 R 2R 3R -2E 18 16 14 12 10 8 3. 6 4 2 R 2R 3R -2 E 18 16 14 12 10 8 4. 2 R 2R 3R -2 009 10.0 points A large flat sheet of charge has a charge per unit area of 9.7 AC/m-. Find the electric field intensity just above the surface of the sheet, measured from its midpoint. The permittivity of free space is 8.8542 x 10 * C-/N . m . Answer in units of N/C. 010 10.0 points Two large, parallel, insulating plates are charged uniformly with the same positive areal charge density to, which is the charge per unit area. What is the magnitude of the resultant elec- tric field E? The permittivity of free space4mke 1. - between the plates, zero outside ED 20 2. Zero between the plates, = CO outside 3. - between the plates, zero outside 2 co 20 4. between the plates, outside 20 5. between the plates, zero outside 6. Zero between the plates, - outside 7. Zero everywhere 8. - between the plates, - outside 2 en 2 en 9. - everywhere 10. Zero between the plates, 2 co outside 011 10.0 points A net positive charge @ is placed on a large, thin conducting plate of area A. The area of the edges of the plate is completely negligible by comparison. In electric equilibrium, find the charge den- sity o on each surface and the magnitude of the electric field [ E] outside the plate. 1. 0 = A and |E] = 0 2. 7 = 2 A and E| = Q 2 A and E= 2 EQ Q 4. = 2 A and JE| =0 5.0= A and |E| = 5 6. 0= and |E| = 2 En 012 10.0 points The difference in potential between the accel- erating plates of a TV set is about 22 kV. Thedistance between these plates is 1.9 cm. Find the magnitude of the uniform electric field in this region. Answer in units of N/C. 013 10.0 points An electron transfers across a potential differ- ence of 8.3 x 105 V. What energy must it be given to accomplish this? Answer in units of J. 014 10.0 points Through what potential difference would an electron need to be accelerated for it to achieve a speed of 2.9% of the speed of light (2.998 x 10* m/s), starting from rest? The mass of an electron is 9.109 x 10 * kg and the elemental charge is 1.602 x 10 C. Ig- nore special relativity. Answer in units of V. 015 (part 1 of 2) 10.0 points Three point charges -9, -2q and q are located at the corners of an equilateral triangle of side -2g P Remove -q to infinity. What is the work done (against the electric force due to 2g and 4) in bringing in charge -q from the infinity to the point bottom right vertex of the triangle? 1. U = 2k 2. U =-3k 3. U = 04. U = k 5. U = -2k 6. U = 4k 7. U = 3k B. U = -k 016 (part 2 of 2) 10.0 points What is the total electric energy stored in the triangular system? 1. U =ky 2. U = -2k 3. U = -3k 4. U = 4k, 5. U = 2k 6. U = 3k 7. U=-k= B. U =0