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 < Fb < Fc Fb < Fc < Fa Fc < Fa < Fb

Fa < Fc < Fb Fc < Fa = Fb 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.

(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

Ftot F23

F12

d=√2 a

q2

q3

q1

a

a

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 y-axis 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 ρa = ρb > ρc > ρd All tie

ρc = ρd > ρa = ρb ρd > ρc > ρb > 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 Ea = Eb > Ec > Ed All tie

Ec = Ed > Ea = Eb 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.

E

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

(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