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PHY2049 Spring 2011 Profs. P. Avery, S. Hershfield
Exam 1 Solution
- A solid conducting sphere of 15 cm radius is located at the center of a spherical conducting shell, whose inner radius is 21 cm and outer radius 42 cm. At a distance of 50 cm from the center of these concentric conductors, the electric field is
- 1 × 105 V/m and is pointing outward. The shell has a net charge of +3. 1 μC. What is the net charge on the sphere at the center in μC?
Answer: +2. 7
Solution: A spherical Gaussian surface of radius 50 cm encloses a charge of 4π(0. 5 m)^2 Eǫo = 5. 8 μC. Since the shell has a net charge of 3. 1 μC, the sphere has a net charge of 5. 8 − 3 .1 = 2. 7 μC.
- A long, straight wire is uniformly charged with a linear charge density of 5.0 μC/m. The wire runs along the axis of a cylinder of radius 2.0 cm and length 30.0 cm. What is the total electric flux through the cylinder surfaces in N·m^2 /C?
Answer: 1. 69 × 105
Solution: From Gauss’s Law, the electric flux is equal to the charge enclosed divided by ǫo. The charge enclosed is (5μC/m)(0. 3 m) = 1. 5 μC so the total electric flux is (1. 5 μC)/ǫo.
- The figure shows a uniformly charged, nonconducting spherical shell of inner radius a and outer radius 2a. If the electric field at the outer radius is E, what is the electric field at point P with radius r = 1. 5 a?
2 a
a
P
Answer: 0. 6 E
Solution: Here we apply Gauss’s law twice: once for a spherical surface of radius 2a and once for a spherical surface of radius 1. 5 a. Let the magnitude of the electric field at 2a be E, and the magnitude of the electric field at 1. 5 a be E′. The charge density is denoted by ρ. ∮ E^ ~ · ˆndA = 4π(2a)^2 E = Qenc/ǫo = ρ(4π/3) ((2a)^3 − a^3 )^ /ǫo
∮ E^ ~ · ˆndA = 4π(1. 5 a)^2 E′^ = Qenc/ǫo = ρ(4π/3)
(1. 5 a)^3 − a^3
/ǫo
Divide the second equation by the first: (1.5)^2 E′ 22 E
(1.5)^3 − 1
which implies that E′^ = 0. 603 E.
- A thick conducting spherical shell has inner radius r and outer radius R, as shown in the figure. A point charge of −Q is located at the center of the sphere and a charge of +q is placed on the conducting shell. The charge on the outer surface of the conducting shell is:
r R
+q
Answer: q − Q
Solution: A Gaussian surface just larger than R will enclose a charge of q − Q, while a Gaussian surface just smaller than R will enclose no net charge because the electric field is zero inside a conductor. Consequently, the charge on the outer surface of the shell must be q − Q.
- Suppose we have an insulating spherical ball of uniform charge density ρ and radius R. At what radius or radii from the center of the sphere is the electric field strength reduced by a factor of 4 from the electric field strength at the surface?
Answer: R/4 and 2R
Solution: Applying Gauss’s law, the flux through a sphere of radius r < R is
Φ = ( E~ · nˆ)4πr^2 =
Qenc ǫo
ǫo
4 π 3
r^3 ρ ⇒ E~ · nˆ =
3 ǫo
rρ.
The flux through a sphere of radius r > R is
Φ = (E ~ · ˆn)4πr^2 =
Qenc ǫo
ǫo
4 π 3
R^3 ρ ⇒ E~ · ˆn =
3 ǫo
R^3
r^2
ρ.
The electric field at the surface, r = R, is E~ · ˆn = ρR/(3ǫo). To determine where the electric field is 1/4 of the value at the surface, set E~ · ˆn for the r < R and r > R cases equal to (1/4)ρR/(3ǫo) and solve for r.
- Two lead nuclei each with charge +82e approach one another head-on from a great distance. Initially each has kinetic energy 50.0 MeV (1 MeV = 1. 6 × 10 −^13 J). What is the closest distance in fm (1 fm = 10−^15 m) the nuclei will approach one another?
Answer: 97
Solution: Initially since the nuclei are far apart their potential energy is zero, Ui = 0. Their initial kinetic energy is 2 ∗ 50 × 1. 6 × 10 −^13 J. The factor of two is because both nuclei have kinetic energy. Finally, their kinetic energy is zero, and their potential energy is k(82e)^2 /r. Use conservation of energy, Ki + Ui = Kf + Uf , to solve for r = k(82e)^2 /Ki.
- Particles 1, 2 and 3 are on the x-axis at x = 0 cm, 5 cm, and 10 cm, respectively. The charges of particles 1 and 2 are each 5 μC. The electrical potential energy of the three-particle system is 0, where potential energy at infinity is defined to be 0. What is the charge of particle 3 in μC?
Answer: − 3. 3
Solution: The net potential energy of this charge configuration is
U =
k(5μC)(5μC) 5 cm
k(5μC)Q 5 cm
k(5μC)Q 10 cm
where Q is the unknown charge. Setting U = 0, we get the following equation for Q: 0 = (5μC) + 1. 5 Q.
- The rectangle shown in the figure has sides 5 cm and 8 cm. The charges are q 1 = 6μC and q 2 = 9μC. How much work is required to move a charge q 3 = 4. 0 μC from point A to point B?
q 1
q 2
A
B
Answer: +0.81 J
Solution: Compute the initial and final potential energy of the charge configuration. The work is equal to W = −(Uf − Ui).
Ui =
k(6μC)(4μC) 8 cm
k(9μC)(4μC) 5 cm
k(6μC)(9μC) √ 89 cm
Uf =
k(6μC)(4μC) 5 cm
k(9μC)(4μC) 8 cm
k(6μC)(9μC) √ 89 cm Note that two of the terms are the same in the initial and final cases.
- A dipole is allowed to rotate freely like the molecules in water. Which of the following statements is true?
Answer: The dipole will be attracted to both a positive and a negative charge.
Solution: For the case of a positive charge the negative end of the dipole will rotate towards the positive charge, and the dipole will be attracted to the positive charge. Similarly, for the case of a negative charge the positive end of the dipole will rotate towards the negative charge, and the dipole will be attracted to the negative charge.
- Two charges with charge Q < 0 are separated by 3 cm. There is a 1 N repulsive force between the charges. How many excess electrons are needed to make one of the charges, Q?
Answer: 2 × 1012
Solution: The magnitude of the force is 1N = kQ^2 /(3cm)^2. This implies that the charge is Q = − 3. 16 × 10 −^7 C, and the number of excess electrons is |Q|/e = 2 × 1012.
- Four charges are placed on the corners of a square of side 2 m as shown in the figure. If Q 1 = 1μC, Q 2 = 3μC, and Q 3 = Q 4 = 2μC, what is the magnitude of the electric field at point P at the center of the square? Answer: 6400 N/C
Solution: The electric fields due to Q 1 and Q 4 partially cancel, as do the electric fields due to Q 2 and Q 3. Add the remaining electric fields as vectors:
Ex Ey E^ ~ 1 + E~ 4 −k(1μC)(
2 m)^2 cos(45) +k(1μC)(
2 m)^2 sin(45) E^ ~ 2 + E~ 3 −k(1μC)(
2 m)^2 cos(45) −k(1μC)(
2 m)^2 sin(45) E^ ~ total − 2 k(1μC)(
2 m)^2 cos(45) 0
- The electric field lines of two point charges, A and B, are shown in the figure. Which of the following is true of the point charges? Answer: Both point charges are positive.
Solution: Electric field lines go away from positive charges and towards negative charges. The electric field lines are going away from both A and B.
- A line of charge in the form of an arc in a circle of radius 2 m is centered about the origin as shown in the figure. If the charge per unit length on the line is λ = 5nC/m, what is the electric field at the origin?
Answer: +22. 5 V /mˆi − 22. 5 V /m ˆj
Solution: Let θ be the usual angle measured with respect to the positive x-axis. The arc extends from θ = 0 to θ = 3π/2. By symmetry the x and y-components of the electric field at the origin will be the same magnitude. Thus, we only need to compute Ex.
Ex = −
∫ (^3) π/ 2
0
kλrdθ r^2
cos θ = −
kλ r
sin(θ)
3 π/ 2 0
kλ r
Ey = −Ex because the electric field goes away from positive charges.
- A line of charge of length L and charge per unit length, λ, is located on the x-axis as shown in the figure. Which expression below gives the x-component of the electric field, Ex, at a distance D from the left end of the line?
Answer: −
∫ L
0
kxλ (x^2 + D^2 )^3 /^2
dx
Solution: The magnitude of the electric field due to an element of charge dq = λdx at x is kdq/r^2 , where r =
x^2 + D^2. To get the x-component of the electric field we must multiply by the trigonometric factor −x/r, where the minus sign is because the electric field due to dq is in the negative x-direction. Thus, the integrand is (kdq/r^2 )(−x/r).
- A proton is located at x = 0 and an electron is located at x = 2 m. There is a uniform electric field of 5V /mˆi. If at t = 0 they are released from rest, at what time would they reach the same x-position? (Do not include the attraction between the proton and the electron.)
Answer: 2. 1 μs
Solution: The proton accelerates to the right (+ˆi) with acceleration ap = eE/mp, and the electron accelerates to the left with ae = eE/me. They pass each other when their positions are the same: xp = 0. 5 apt^2 = xe = 2m − 0. 5 aet^2 , which implies a time of t =
4 m/(ap + ae).
