September 5, 2008
PHY2054 Discussion-Fall ‘08
Quiz 1 (Chapter 15.1-15.6)
Name:
UFID:
*1. (5pts) A glass object is charged to +6.00 nC by rubbing it with a silk cloth. In the rubbing process,
how many of the electrons are removed from it? (Electronic charge -e = -1.60×10-19 C)
The total charge of electrons removed from the glass is -6 nC. Since each electron has the charge of
-1.60×10-19 C, the number of electrons is given by
n = -6×10-9/-1.60×10-19 C = 3.75×1010
**2. (5pts) Three point charges lie along the x-axis. The negative charge q1 = -2 μC is at the origin
and the positive charge q2 = +6 μC is at x = +2.00 m. Where must a negative charge q3 be placed on
the x-axis so that the resultant electric force on it is zero?
We take +x axis to the right. For the net force to be zero, q3 must be nearer to q1 than to q2 because
|q2| is larger than |q1|. Also, q3 should be left of the q1 since the electric forces exerted by q1 and q2
have to be opposite in directions. Thus x<0. Applying the equilibrium conditionΣF = 0, we have
0 = F1+F2 = -k|q1||q3|/|x|²+k|q2||q3|/(|x|+2)² = k|q3|[-|q1|/x²+|q2|/(-x+2)²]
⇒ -2(-x+2)²+6x² = 0 ⇒ 4x²+8x-8 = 0 ⇒x²+2x-2 = 0
⇒ x = [-2-√(2²+4×2)]/2 = -2.73 m. 2.73 m left of the origin.
**3. (5pts) Protons are projected with an initial speed of v0 = 1.00×104 m/s into a region where a
uniform downward electric field with intensity E = 850. N/C is present. The protons are to hit the
target that lies a horizontal distance of 1.00 mm from the point where the protons are launched. Find
the two projection angles that result in a hit. (Mass of a proton mp = 1.67×10-27 kg)
The acceleration of the protons is given by
mpa = qE ⇒ a = qE/mp = 1.6×10-19×850/1.67×10-27 = 8.14×1010 m/s²
When the protons hit the target, the vertical displacement is zero. The flight time is expressed as
0 = v0sinθt-(1/2)at² = t(v0sinθ-(1/2)at) ⇒ t = 2v0sinθ/a
Using the kinematics equation for the horizontal displacement, we have
Δx = v0cosθt = v0cosθ[2v0sinθ/a] = v02sin(2θ)/a
sin(2θ) = aΔx/ v02 = 8.14×1010 ×1×10-3/(1.00×104)² = 0.814
2θ = sin-1(0.814) = 54.5˚, 125.5˚ (Supplementary angle)
Thus, θ = 27.25˚, 62.25˚
***4. (5pts) A 2.00-g sphere is suspended as a pendulum by a 0.500-m light string in the presence of
a uniform electric field. The field has an intensity of 8×105 N/C and is directed at an angle of 30.0˚
to the horizontal. What is the period of the simple harmonic motion of the pendulum if the sphere
has a charge of +9.00 nC? (Hint: The period of the pendulum is given by T = 2π√(L/geff), where geff
is effective gravitational acceleration, which satisfies mgeff = tension at the equilibrium position.)
Three forces are exerted on the pendulum: gravity mg, tension F and electric force qE. Applying the
equilibrium condition, we have
ΣFx = 0 ⇒ qEcos30˚-Fsinθ = 0 ⇒Fsinθ = qEcos30˚
ΣFy = 0 ⇒ qEsin30˚+Fcosθ-mg = 0 ⇒ Fcosθ = mg-qE sin30˚
where the θ is the angle the string makes to the vertical. Using a trigonometric identity cos²θ+sin²θ
= 0 we solve the simultaneous equation for F.
F² = (qEcos30˚)²+( mg-qE sin30˚)² ⇒ F = √[(qEcos30˚)²+( mg-qE sin30˚)²]
The effective gravitational acceleration is given by
geff = F/m = √[(qEcos30˚/m)²+( g-qE sin30˚/m)²] = √[(3.6×cos30˚)²+(9.8-3.6sin30˚)²] = 8.59 m/s²
Thus the period is
T = 2π√(L/geff) = 2×3.14√(0.5/8.59) = 1.52s