January 20, 2009
PHY2054 Discussion-Spring ‘09
Quiz 1 (Chapter 15.1-15.9)
Name:
UFID:
**1. (2.5pts) Two 10.0-g spheres are suspended by 25.0-cm strings as pendulums. The two
pendulums are separated by 5.00 cm at the ceiling. The spheres are given the same electric charge.
Find the magnitude of the charge on each sphere if each string makes 15.0º to the vertical when they
are in equilibrium.
The distance between two charged spheres is
r = 2Lsinθ+d = 2×0.25×sin15º+0.05 = 0.179 m
Since both spheres are in equilibrium, the net force on each is zero.
Applying Newton’s 2nd law to the vertical and horizontal direction, we
have
ΣFy = 0 ⇒ Tcosθ-mg = 0 ⇒ T = mg/cosθ
ΣFx = 0 ⇒ Fc-Tsinθ = 0 ⇒ Fc = Tsinθ = (mg/cosθ)sinθ = mgtanθ
Using Coulomb’s law we express Coulomb force in terms of charge and distance, and solve the
equation for the charge.
kq2/r2 = mgtanθ ⇒ q = √(mgtanθ/k)×r = √(0.01×9.8×tan15º/8.99×109)×0.179 = 3.06×10-7 C
**2. (2.5pts) Three point charges are located at the corners of an equilateral triangle with sides of
length 0.500 m. 2.00-μC is located at the origin, -4.00μC on the +x axis and 7.00-μC at the other
corner. Calculate the net electric force on the -4.00-μC charge.
First we calculate the magnitudes of electric forces on -4-μC (≡ q3) charge
due to 2-μC (≡q1) and 7-μC (≡q2) charges respectively.
F13 = k|q1q3|/r2 = 8.99×109×(2×10-6)×(4×10-6)/0.52 = 2.88×10-1 N
F23 = k|q2q3|/r2 = 8.99×109×(7×10-6)×(4×10-6)/0.52 = 10.07×10-1 N
To find the net force, we express each force in components and then add them
up.
F13 = (F13cos(180º), F13sin(180º)) = (-2.88×10-1 N, 0 N)
F23 = (F23cos(120º), F23sin(120º)) = (-5.04×10-1 N, 8.72×10-1 N)
F3 = (F13x+F23x, F13y+F23y) = (-7.92×10-1 N, 8.72×10-1 N)
The magnitude and direction of the force are calculated as
F3 = √(F3x2+F3y2) = 1.18 N
θ = tan-1(F3y/F3x) = -47.8+180 = 132.2º
Constants & Formulas
k = 8.99×109 Nm²/C²
me = 9.11×10-31 kg
Coulomb Force: Fe = k|q1||q2|/r2
ε0 = 8.85×10-12 C²/(Nm²)
mp = 1.67×10-27 kg
Electric Field: E = Fe/q0
Gauss’s Law: Ф = Qin/ε0
e = 1.6×10-19 C
Electric Flux: Ф = EAcosθ
**3. (2.5pts) Two point charges are placed along the x-axis. -4.0 μC is at the origin, 8.00 μC is 2.00
m from the origin. Locate the point at which the net electric field is zero.
The electric fields due to the two charges are in the same direction
between the charges, and the electric field due to 8-μC charge is
larger than that due to 4-μC in the region to the right of 8-μC
charge. Therefore, if there is a position where the net field is zero, the position is to the left of -4-μC
charge. In this region, the distance from the 8-μC charge is |x|+2 = -x+2 m, thus we have
E1-E2 = 0 ⇒= k|q1|/x2- k|q2|/(-x+2)2 = 0 ⇒ 4(-x+2)2-8x2 = 0 ⇒ x2+4x-4 = 0
⇒ x = [-4±√(42+4×4)]/2 = -2±2√2
Since x < 0, only the negative root gives a proper solution.
x = -2-√2 = -4.83 m, 4.83 m to the left of the -4-μC charge.
*4. (2.5pts) A conducting sphere is located at the center of a conducting spherical shell. The sphere
carries a net charge of -12.0 μC and the shell a net charge of 7.00 μC. What is the net charge on the
outer surface of the spherical shell?
We choose the Gaussian surface as a sphere inside the conducting shell. Since the electric field
inside a conductor is zero, we have
Ф = Qin/ε0 ⇒ 0 = (qsphere+qshell, in)/ε0 ⇒qshell, in = -qsphere= +12.0 μC
The net charge on the shell is 7 μC, thus the charge on the outer surface is
qshell, in+qshell, out = qshell ⇒ qshell, out = qshell,-qshell, in = 7-12 = -5.00 μC