Midterm Practice 2
Physics 1B ------------------------------------ Name: ------------------------------------ ID number: ------------------------------------ Lecture (1 or 2): Section (A-F): ------------------------------------
1. A nonconducting spherical shell, with an inner radius R1 = 4 cm and an outer radius R2 = 6 cm, has charge spread non-uniformly through its volume between its inner and outer surfaces. The volume charge density is the charge per unit volume measured in coulombs per cubic meter. For this shell, b (r) = , r where r is the distance to the center, and b = 3 10-6 C/m2 . What is the total net charge in the shell? Solution.
R2 R1
Q=
dV =
2 2 (r)(4r 2 ) dr = 2b R2 - R1 = 3.8 10-8 C
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2. Two particles are fixed to an x axis: particle 1 of charge q1 = 2.1 10-8 C at x1 = 20 cm and particle 2 of charge q2 = -4q1 at x2 = 70 cm. At what coordinate on the axis is the net electric field produced by the particles equal to zero? Solution. The charges have opposite sign, so the field can only cancel to the left of both of them or to the right of both of them (not between the two). The total field is q -4q (x - x1 )2 +k =0 = 4 x = -30 cm (x - x1 )2 (x - x2 )2 (x - x2 )2
E=k
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3. An electric dipole consisting of charges of magnitude Q =1.5 nC separated by l =6.2m is in an electric field of strength |E| = 1100 N/C. What are (a) the magnitude of the electric dipole moment and (b) the difference between the potential energies for the dipole orientations parallel and antiparallel to E? Solution. (a) The dipole moment is p = Ql = 9.3 10-15 Cm. (b) To flip the orientation of the dipole, one has to move one charge around other, the i.e. move it in the electric field by distance 2l. The change in the potential energy is U = Q E (2l) = 2.05 10-11 J (One could also show that, in general, the energy of the dipole p in the electric field E is U = p E. This leads to the same answer as above.)
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4. A charge distribution that is spherically symmetric but not uniform produces an electric field of magnitude |E| = E(R) = bR4 , directed radially outward from the center of the sphere. Here R is the distance from the center, and b is some constant. What is the volume density of the charge distribution? Solution. Using the spherical symmetry, one can relate the flux through a spherical surface at distance R to the electric field at this surface: = 4R2 E(R) = 4bR6 According to Gauss' Law the same flux is related to the charge contained inside this surface:
R
= 4k
dV = 4k
0
(r)(4r 2 )dr,
where k = 1/(4 0 ). Now let us differentiate both sides of the equation
R
4bR6 = 4k
0
(r)(4r 2 )dr
with respect to R. This yields the final result (R) = 6b 0 R3
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5. Two electrons are fixed at distance d = 2 cm apart. The charge of the electron is e = 1.6 10-19 C, and its mass is me = 9.1 10-31 kg. Another electron is shot from infinity and stops midway between the two. What is its initial speed? Solution. Assuming the potential is zero at infinity, the potential at the point between the two electrons is V = 2k 4ke e = , d/2 d
The potential energy of the third electron at this point is U = eV = 4ke2 d
Since the electron is at rest, its initial kinetic energy me v 2 /2 has been converted to the potential energy: mv 2 4ke2 = v= 2 d 8ke2 = 320 m/s me d
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