Download Louisiana State University Physics 2102, Exam 1

Name: ______________ ID No. : ______________ Instructor:____Key__________
Louisiana State University Physics 2102, Exam 1,
September 22, 2004
• Please be sure to write your name, student ID number and class instructor above.
• For the problems: Show your reasoning and your work.
• The questions are multiple choice. No partial credit on the questions.
• You may use scientific or graphing calculators, but you must derive and explain your
answer fully on paper so we can grade your work.
• Feel free to detach, use, and keep the formula sheet pages. No other reference material
is allowed during the exam.
• Good Luck!
Question 1 [8 points]
The figure shows three arrangements of an electron and two protons.
(i) (3 pts) Draw the forces on the electron due to the two protons, ine ach arrangement.
(ii) (5pts) Rank the arrangements according to the magnitude of the net electrostatic force
on the electrons due to the protons, smallest first. Circle the right answer.
Fa < F b < Fc
Fa < Fc < F b
Fb < Fc < F a
Fc < Fa < F b
Fc < Fa = F b
All tie
Problem 1 [17 points]
Three point charges, q1 = -1.2 x 10-8 C, q2 = -2.6 x 10-8 C and q3 = +3.4 x 10-8 C, are held
at the positions shown in the figure, where a = 0.16 m.
q3
d=√2 a
a
F23
Ftot
F12
q1
a
q2
(a) (3 pts) Draw the forces acting on q2 (including the total force, approximately)
(b) (5 pts) Calculate the x-component of the total force acting on q2 due to the other two
charges.
|F12| = kq1q2/a2 = 8.99 109 Nm2/C2 x(1.2 10-8 C)x(2.6 10-8 C)/(0.16m)2 = 1.10 10-4 N
F12 is in the x-direction, so F12x = +|F12| = +1.096 10-4 N
|F23| = kq2q3/d2 = 8.99 109 Nm2/C2 x(2.6 10-8 C)x(3.4 10-8 C)/(√2 x 0.16m)2 = 1.55 10-4 N
F23x = −|F23|cos 45o = −1.098 10-4 N
Ftot x = F12x+F23x = −1.9 10-7 N ~ 0
(c) (5 pts) Calculate the y-component of the total force acting on q2 due to the other two
charges.
F12y = 0
F23y = +|F23|sin 45o = 1.098 10-4 N
Ftot y = F12y+F23y = 1.098 10-4 N
(d) (4 pts) Calculate the magnitude of the total force acting on q2 .
|Ftot|=(Ftotx 2 + Ftoty 2) ½ = 1.1 10-4 N
Question 2 [8 points]
The figure shows an electron in a uniform (external) electric field.
(a) (3 pts) What is the direction of the electrostatic force on the electron due to the
electric field shown? (Circle the right answer).
(b) (2 pts) In which direction will the electron accelerate if it is moving parallel to the yaxis before it encounters the electric field? (Circle the right answer)
(c) (3 pts) If instead, the electron is initially moving rightward, will its speed increase,
decrease, or remain constant? (Circle the right answer).
Increase
Decrease
Remain the same
Problem 2 [17 points]
The figure shows an electric dipole in a electric field. Use q = 6 nC, d = 4 nm, and θ =
40o. The magnitude of the electric field is 290 kV/m.
(a) (4 pts) Draw the dipole moment on the figure.
|p| = qd, from the negative charge to the positive charge
(b) (4 pts) Calculate the electric potential energy of the electric dipole in the
configuration shown in the figure.
U = −p ·E = - |p| |E| cos θ = − q d E cos θ
= − 6 10-9 C x 4 10-9 m x 290 103 V/m x cos 40o = − 5.33 10 -12 J
(b) (4 pts) What would be the electric potential energy of the electric dipole if it were
turned 180o?
New angle between dipole and electric field: θ = 40o + 180o = 220o
New potential energy:
U = −p ·E = - |p| |E| cos θ = − q d E cos θ
= − 6 10-9 C x 4 10-9 m x 290 103 V/m x cos 220o = + 5.33 10 -12 J
(c) (5 pts) Calculate the work required to turn the electric dipole 180o.
Wext = ∆U = Uf – U0 = + 5.33 10 -12 J – (−5.33 10 -12 J) = 1.07 -11 J
Question 3 [8 points]
The figure shows four spheres, each with charge Q uniformly distributed through its
volume.
(i) (4 pts) Rank the spheres according to their volume charge density ρ, greatest first.
Circle the right answer:
ρ = Q/V, Va<Vb<Vc<Vd, so:
ρa > ρb > ρc > ρd
ρc = ρd > ρ a = ρb
ρa = ρb > ρc > ρd
All tie
ρd > ρc > ρ b > ρa
The figure also shows a point P for each sphere, all at the same distance from the center
of the sphere.
(ii) (4 pts) Rank the spheres according to the magnitude of the electric field they produce
at point P, greatest first. Circle the right answer.
Ea > Eb > Ec > Ed
Ec = Ed > Ea = Eb
Ea = Eb > Ec > Ed
All tie
Ed > Ec > Eb > Ea
Gauss’ law with spherical symmetry says that E(r)=kqenc/r2. For (a) and (b), qenc=Q, so
they tie. For (c), qenc<Q, and for (d), qenc is even smaller.
Problem 3 [17 points]
A square metal sheet with side lengths d = 20 cm is charged with a total charge Q = +15 mC.
(a) (5 pts) Calculate the surface change density σ on each surface of the sheet.
On a conductor, the charge will spread out on both sides. The charge
on each side is q=Q/2, and the charge density on each side is
σ = q/A =(Q/2)/d2 = (15 10-3 C/2)/ (0.2 m)2 = 0.19 C/m2
E
(b) (5 pts) What is the direction and magnitude of the electric field to the left of the sheet,
close to the surface of the sheet?
The electric field points away from positive charges, so it weill point away from the sheet.
Close to the surface of the conductor,
E = σ/ε0 = 0.19 C/m2 / 8.85 10-12 C2/Nm2 = 2.1 1010 N/C
(c) (5 pts) Calculate the electrical force on a point charge q = +2 µC, located 0.2 mm to the
left of the sheet (i.e. close to the surface).
|F| = q |E| = 2 10-6 C x 2.1 1010 N/C = 4.3 104 N
(d) (2 pts) Would the answer to the question in (c) be the same, if the point charge was
located 20 m to the left of the sheet (i.e. far away from the sheet)? (Answer yes or no, and
why).
No: since 20m is much larger than the sheet side (20cm), the sheet would look more like a
point charge, and |F| ~ kqQ/r2 (smaller).
Question 4 [8 points]
The figure shows a family of parallel equipotential surfaces (in cross section), and five
paths along which we shall move an electron from one surface to another.
(a) (4 pts) What is the direction of the electric field associated with these surfaces? Circle
the right answer.
The electric field is perpendicular to the equipotentials, and points towards lower
potential:
(b) (4 pts) Rank the paths according to the work we do, greatest first. Circle the right
answer.
W1 > W2 > W3 > W4 > W5
W1 > W5 > W3 > W4 > W2
W4 > W1= W2 = W5 > W3
All tie
W3 > W1 = W2 = W5 > W4
W1= W2 > W3 > W4 = W5
Wext = ∆V q = -e ∆V : if ∆V is -ve, Wext is +ve. Paths 1, 2 and 5 have ∆V = -10 V and
W1=W2=W3=+10eV; path 3 has ∆V = -20V and W3=+20 eV; path 4 has ∆V=+10V and
W4=-10eV.
Problem 4 [17 points]
The rectangle below has side lengths of 5.0 cm and 15 cm. The charges are q1 = -5.0 µC
and q2 = +2.0 µC. Let V = 0 at infinity.
(a) (5 pts) Calculate the electric potential at point A.
VA= V1 + V2 = k q1/d1+ k q2/ d2 = k(q1/d1+q2/d2)
= 8.99 109 Nm2/C2 (-5 10-6 C/0.15 m + 2 10-6 C/ 0.05m) = 60 kV
(b) (5 pts) Calculate the electric potential at point B.
VB= V1 + V2 = k q1/d2+ k q2/ d1 = k(q1/d2+q2/d1)
= 8.99 109 Nm2/C2 (-5 10-6 C/0.05 m + 2 10-6 C/ 0.15m) = - 779 kV
(c) (7 pts) How much work is required to move a third charge q3 = +3.0 µC from B to A
along a diagonal of the rectangle?
Wext = ∆V q3 = (VA – VB) q3 = (60 kV – (-779 kV)) 3 10-6 C = 2.5 J