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PHY2049 Spring 2010 Profs. P. Avery, A. Rinzler, S. Hershfield
Exam 1 Solution
- Four parallel infinite sheets of charge spaced by 4 cm between adjacent sheets are shown edgewise in the figure. Their surface charge densities are as indicated with σ = 2. 0 μC/m^2. What is the magnitude of the electric field (in N/C) at point P located midway between two sheets as shown? (Hint: Electric fields are vectors that add as such.)
A B C D
P
+! ! 2 "
Answer: 4. 5 × 105
Solution: Sheet A produces an electric field of magnitude σ/(2ǫo) going away from it, which means the electric field is pointing towards the bottom of the page at point P. Sheet B produces an electric field of magnitude 2σ/(2ǫo) going towards it, which implies that the electric field is pointing towards the bottom of the page at point P. Similarly, at point P the electric fields due to sheets C and D are 3σ/(2ǫo) toward the top of the page and 4σ/(2ǫo) toward the bottom of the page. Adding these four fields together produces a net electric field of 4σ/(2ǫo) toward the bottom of the page.
- 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 16 from the electric field strength at the surface?
Answer: R/16 and 4R
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 1/16 of the value at the surface, set E~ · ˆn for the r < R and r > R cases equal to (1/16)ρR/(3ǫo) and solve for r.
- A non-uniform electric field given by E~ = (5.5ˆi − 2 .1ˆj + (4. 6 z^2 − 3)kˆ) N/C pierces a cube with sides 3 m, as shown in the figure. The cube has its rear corner at the origin. What is the total charge inside the cube?
y
z
Answer: +3.3 nC x
Solution: By Gauss’s law the net flux is equal to Qenc/ǫo. The flux through each face of the cube is ( E~ · ˆn)(3 m)^2 , where the normal vector points outwards. For the front face the normal component of the electric field is ( E~ · nˆ) = 5.5 N/C, while for the back face it is ( E~ · nˆ) = − 5 .5 N/C. These two fluxes will cancel in computing the net flux. Similarly, the flux for the right and left faces will cancel. Only the fluxes through the top and bottom faces do not cancel. The flux through the top face (z = 3) is (4. 6 · 32 − 3) 3^2 , and the flux through the bottom face is −(4. 6 · 02 − 3) 3^2. Combine these to find the charge enclosed is ǫo 4. 6 · 34.
- Members of a space mission discover a mysterious alien artifact consisting of a solid conducting sphere of radius 4 cm floating inside a conducting shell with inner and outer radii 10 cm and 12 cm, respectively. Scientists measure the electric field strength at different places, obtaining E = 4500 N/C pointing inward at r = 7 cm and E = 7600 N/C pointing outward at r = 15 cm. What is the total charge on the outer shell?
Answer: +21.5 nC
Solution: For a Gaussian spherical surface of radius 7 cm, the charge enclosed is that of the sphere, while for a Gaussian spherical surface of radius 14 cm, the charge enclosed includes both the sphere and the shell. Apply Gauss’s law,
Φ = ( E~ · nˆ)4πr^2 = −4500(4π 0. 072 ) =
Qsphere ǫo
Φ = ( E~ · ˆn)4πr^2 = 7600(4π 0. 152 ) =
Qsphere + Qshell ǫo
Solve the first equation for Qsphere, and then use that to find Qshell in the second equation.
- A long solid copper cylinder has radius 1 cm and length of 90 cm. The field at the surface of the cylinder is 9000 N/C going outwards. What is the charge on the cylinder?
Answer: 4.5 nC
Solution: Use a cylindrical Gaussian surface on the surface of the copper rod. The electric field near the ends is nearly parallel to the surface, and the surface area of the ends is much smaller than that of the side. Thus, the flux from the ends is much smaller than that from the side, which is equal to Φ = 2π(0. 01 m)(0. 9 m)(9000 N/C). The charge enclosed is ǫoΦ.
- Two isolated conducting spheres are separated by a large distance. Sphere 1 has a radius of R and an initial charge 3Q while sphere 2 has a radius of 3R and an initial charge 7Q. A very thin copper wire is now connected to the spheres. Charge flows between the spheres until they reach the same electrical potential. How much charge will be transferred from sphere 2 to sphere 1? (Note that the charge transferred can be positive, negative or zero .)
Answer: −Q/ 2
Solution: The net amount of charge on both spheres does not change because charge is conserved. Thus, in the end q 1 + q 2 = 10Q. Also, in the end they have the same potential so kq 1 /R = kq 2 /(3R), which implies that q 2 = 3q 1. Solve these two equations to find in the end q 1 = 2. 5 Q, and q 2 = 7. 5 Q. The charge on sphere 1 has changed by −Q/2, which means that −Q/2 flowed to sphere 1 from sphere 2.
- A plastic rod has been bent into a circle of radius R = 5.5 cm. It has a charge q 1 = +4.2 pC uniformly distributed along three-quarter of its circumference and a charge q 2 = − 4 q 1 uniformly distributed along the rest of the circumfer- ence (see figure). Assuming V = 0 at infinity, what is the electric potential in volts at point P, which is on the central axis of the circle at distance D = 6.5 cm
from the center? q
q 2 1
P
R
Answer: − 1. 33
Solution: All points on the circumference are r =
R^2 + D^2 from point P. Thus, the potential at point P is
V =
kdq r
k r
dq =
k r
(q 1 + q 2 ).
- In the figure, the central particle of charge 4.0 nC is surrounded by the square array of charged particles (also in units of nC). The square has side length of 1 mm. The magnitude of the net force on the central particle due to the others is (Hint: if you don’t use symmetry to greatly reduce the work of solving this problem come back to it last).
Answer: 0.29 N
Solution: The forces due to charges opposite one another on the square cancel because they have the same magnitude, but opposite directions. The only charge without another charge opposite it is the 2 nC charge on the left. Thus, the magnitude of the net force on the central particle is k(2 nC)(4 nC)/(0. 0005 m)^2.
- In the figure the gray rod has uniform linear charge density λ. The z com- ponent of the infinitesimal electric field at point zp due to the infinitesimal element of charge contained in the infinitesimal line element dx can be written dEz =
Answer: k
zpλdx (x^2 + z^2 p )^3 /^2
Solution: The distance from a charge segment dq = λdx to point p is r =
x^2 + z^2 p. A charge segment produces an electric field of magnitude kdq/r^2 at point p. To get the z-component of the electric field one must multiply by the trigonometric factor zp/r. (As a check note that when x = 0, r = zp and zp/r = 1.) Putting the magnitude and the trigonometric factor together, we obtain dEz = (kdq/r^2 )(zp/r) = kλdx(zp/r^3 ). For the choices given on the exam, it also turns out that only the above answer has the correct units of the electric field. Thus, just by checking units, one could have gotten this problem correct.
- The figure shows 3 insulating hollow spheres that have the indicated charge (aQ) uniformly distributed on their surfaces. In each case there is a point particle of the indicated charge (bq) a distance d from the center of the hollow sphere. Rank order the electrostatic force between the sphere and the point charge for the three cases.
Answer: C > B > A
Solution: In case A the force is zero because the charge is inside the hollow sphere. For cases B and C the forces are k(4Q)(2q)/d^2 and k(6q)(3Q)/d^2 , respectively.
- An electron traveling along the x axis (toward increasing x) has a speed of 4. 0 × 106 m/s. At time t = 0 it enters a region
of space in which there is an electric field of 10.0 N/Cˆj. At a time t = 10μs later the electron’s velocity makes an angle with respect to the +x axis that is:
Answer: − 77 ◦
Solution: The force on the electron in the electric field is −|e|(10 N/C)ˆj, and consequently its acceleration is a = −|e|(10 N/C)/meˆj. This problem is thus similar to a projectile motion problem from Physics 1. At a time t later the velocity in the x and y directions is vx = 4 × 106 m/s and vy = 0 − |a|t. The angle the velocity vector makes with the x-axis is tan−^1 (vy /vx).
- The figure shows the electric field lines in three different regions of space. The points labeled R and S are the same distance apart in each region. A proton is let go from rest at point R in each region. Rank order the proton momentum on reaching S for the three cases.
Answer: A > B > C
Solution: When the electric field lines are closer together, the electric field is larger. Thus, the electric field in case A is larger than that in case C. In case B the electric field is not going uniformly to the right as in cases A and C. However, between R and S the field is going uniformly to the right with magnitude in between that of cases A and C because the separation of field lines is intermediate between cases A and C. The final momentum of the proton is proportional to the force on the proton, which in turn is proportional to the electric field between R and S. From the above argument the electric fields in the three cases satisfy EA > EB > EC.
- A semi-circular, thin, insulating rod with a total charge of +80.0 nC uniformly distributed along its length, has a radius of 10 cm. If we needed to express an infinitesimal element of charge on the rod, dq, in terms of the infinitesimal angular variable dθ measured in radians, such that dq = udθ, (as we would if we wanted to determine the electric field at the origin, see figure) then for this case u is:
Answer: 25 nC/radian
Solution: The net charge on the rod is 80 nC, and the angle subtended in radians is π (half of a circle). Consequently, the constant u = dq/dθ is equal to 80nC/π.
- Two point particles, with charges of q 1 and q 2 , are placed a distance r apart. The electric field is zero at a point P between the particles on the line segment connecting them. We conclude that:
Answer: q 1 and q 2 must have the same sign but may have different magnitudes.
Solution: In order for the electric fields to cancel the fields due to q 1 and q 2 must be in opposite directions. Since point P is in between the charges, q 1 and q 2 must have the same sign in order for the fields to be in opposite directions. It is not necessary for the charges to be of the same magnitude because point P could be closer to one of the charges than the other.
- A line of charge extending along the x axis has a continuous charge density given by λ = (5x) nC/m. How much charge is contained on the line from x = 2 to x = 4 meters?
Answer: 30 nC
Solution: The charge between 2 and 4 meters is
q =
dq =
2
λdx =
x^2
4 2
nC.
