I don\'t know how to solve this 3. A charge +q arranged as a solid, uniform sphe
ID: 1869382 • Letter: I
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I don't know how to solve this
3. A charge +q arranged as a solid, uniform sphere of radius a is placed at the center of a spherical conducting shell of inner radius b and outer radius c. The shell carries a net charge of +2q. a = 3.00 cm, b= 4.00cm, c= 4.50cm, q=1.00 PC. Find the magnitude of the electric field at A. (9 pts.) r= 2.0 cm: Answer: 6.67 N/C B. (6 pts.) r = 3.5 cm: Answer: 7.35 N/c C. (6 pts.)r=5.0 cm: Answer: 10.8 N/C D. (4pts) How much cliarge appears on the outer surface of the conducting shell? Answer: 3 pc 4. An electron is fired with speed v,=9,0x 106 m/s toward a pair of large, parallel planes of uniform charge per unit area, to, as shown. The electron travels along a line toward a small hole at the center of each plane. The planes are separated by a distance d=1 mm. 2d * EFE. O = 80 pc/m2 A. (10 pts.) What is the magnitude of the electric field at a point between the planes, a distance of d/4 from the positive plane? Answer: 9.04 x 10°N/C B. (15pts.) What is the speed of the electron when it reaches the second plane at point b? Answer: 5.7x10'm/s.Explanation / Answer
In case of charge distributed over concentric spheres uniformly,
electric field at any point, r distance from the common center is given by
E = k q / r2
where k is constant, 9x109
q is the charge inside the sphere of radius r
A) Charge inside sphere of radius 2 cm = 1*23/a3 ( cahrge is propotional to volume, and volume is proportional to r3)
= 8/27 pC
E = 9x109 * 8* 10-12/27* 0.022
= 6.67 N/C
B) Charge inside sphere of radius 3.5 cm = 1pC
E = 9x109*1x10-12/ 0.0352 = 7.35 N/C
C) Charge inside sphere pf radius 5 cm = 3q = 3pC
E = 9x109 * 3* 10-12/0.052
= 10.8 N/C
D) Electric field inside the conductors is zero. So field from r= 4 cm to r =4.5 cm is zero
so charge on inner surface of the shell = -1pC ( to make net charge inside the shell =0)
Total charge on the shell = +2 pC
Hence charge on the outre surface = +3pC
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