Name: Final Exa n-conducting shell of inner radius r dis the electric field at a

Question

Name: Final Exa n-conducting shell of inner radius r dis the electric field at a distance 10 cm and perical, Outer (5 points) Aa total charge 15 C distributed unif (5 poin5? carriesa total the magnitude of the elect uted uniformly througer radius 12 cr itse 0 8 cm carries a the volume of the shelW Ra o the shell. What from the oenter of the shell? A. Zero B. 2.87 x 10 N/C C. 5.75 x 1O N/C D. 2.87 x 10 N/C E. 575 x 10 N/C m carrying positive charge q is sliding down a ram 9. (5 points) A block of mass m the diagram sh kec filetpam shown As the box accelerates downward, find the terninal p with There is a uniform magnetic held of strength B pointing of kinetic fr coeficient out of the page in speed of the block. nward, find the terminal (sine - cos ) B. (sin +J cos 0) (cose-n sin ?) r, 4 AL 10. (5 points) A charged particle is moving with speed 2n dicular to a uniform mag- entical charged particle is moving with netic field. A second id to the same m frequency of revolution of the second particle is speed 2v perpendicular agnetic field. If the frequency of revolution of the first particle is f, the A. f/4 B. f/2 D. 2f E. 4f 11. (5 points) A 6.0?F capacitor is connected in series with a 5.0 M10 resistor, and this combination is connected across an ideal 15 V DC battery. What is the current in the circuit when the capacitor has reached 20% of its maximum charge? A. 1.3? B. 2.4 pA C. 4.7 HA D. 6.51A E.9.1 ?? Physics 230 Spring 201 Mour 23 2018 Page 6 of 7

Explanation / Answer

Solution:

The volume of the shell is (4/3)?(15³-10³) = 9948cm³

The charge density = 15 ?C/9948cm³ = 1.508x10?? C/cm³

The volume of charge inside r=12cm is (4/3)?(12³-10³) = 3049cm³

The amount of charge inside r=12cm = 3049 x 1.508x10?? = 4.60x10?? C

From Gauss's Law, the field at 12cm (=0.12m) is due only to the charge inside r=12cm:

E = kq/r²
= 8.99x10? x 4.60x10??/0.12²
= 2.87x10? N/C [D].

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