January 27, 2009
PHY2054 Discussion-Spring ‘09
Quiz 2 (Chapter 16.1-16.8)
Name:
UFID:
**1. (2.5pts) On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is
also a strong vertical electric field that is uniform close to the planet's surface. A 1.00-kg ball with a
charge of -5.00×10-4 C is thrown straight up at a speed of 18.0 m/s from the ground. It hits the
ground 3.00 s later. What is the potential difference between the starting point and the top point of
the trajectory? (Calculate Vtop-Vbottom.)
When the ball hits the ground, the vertical displacement is zero. Thus we have
Δy = v0t+(1/2)at2 ⇒ 0 = v0t+(1/2)at2 ⇒ a = -2 v0/t = -2×18/3 = -12.0 m/s2
Using Newton’s second law, we calculate the electric field.
ma = qE-mg ⇒ E = (g+a)/q = (9.8-12)/(-5×10-4) = 4.40×103 N/C
The velocity of the ball zero at the peak of the motion. Therefore, the maximum height is
v2 = v02+2aΔy ⇒ 0 = v02+2aΔy ⇒ Δy = - v02/2a = -182/2×(-12) = 13.5 m.
The potential difference is
ΔV = -EyΔy = -4.4×103×13.5 = -5.94×104 V
*2. (2.5pts) Three charges are situated at corners of a rectangles shown as below. How much energy
would be expended in moving 3.00-μC charge to infinity?
The expended energy is the same as the initial potential energy of
3-μC charge because the electric potential energy is zero at infinity.
Thus we have
PEi = qVi = qk(q1/r1+q2/r2)
= 3×10-6×9×109×(8×10×10-6/0.05-2×10-6/0.04) = 2.97 J
Constants & Formulas
Electric potential: ΔV = ΔPE/q
Capacitance: C = Q/ΔV
Potential in a uniform field: ΔV = -ExΔx
Capacitors in parallel: Ceq = C1+C2
Potential due to point charges: V = kq/r
Capacitors in series: 1/Ceq = 1/C1+1/C2
g = 9.80 m/s2
k = 8.99×109 Nm²/C²
**3. (2.5pts) The combination of four capacitors below is connected across 20.0-V battery. What is
the potential difference across the 6.00-μF capacitor?
The 12-μF capacitor and the 4-μF capacitor are connected in series.
1/C’= 1/C1+1/C2 = 1/12+1/4 = 1/3 ⇒ C’ = 3.00 μF
C’ is connected in parallel with 6-μF capacitor.
C” = C’+C3 = 3+6 = 9.00 μF
C” is connected in series with the 18-μF capacitor. Thus the equivalent
capacitance of the combination is
1/Ceq = 1/C”+1/C4 = 1/9+1/18 = 1/6 ⇒ Ceq = 6.00 μF
Since C” and C4 are connected in series they store the same amount of charge, and the charge also
equals to the total charge on the equivalent capacitance.
Q” = Qtot = CeqΔV = 6×10-6×20 = 120. μC
C’ and 6-μF capacitor are in parallel, thus the potential drop across them are equal. We have
ΔV3 = ΔV” = Q”/C” = 120×10-6/9×10-6 = 13.3 V
***4. (2.5pts) Three capacitors and two switches are connected across a battery as shown below.
First switch 2 is closed while switch 1 remains open. After the capacitors are fully charged, switch 2
is opened, and then switch 1 is closed. Find the charge on 2.00-μF capacitor.
Initially the switch 1 is open, thus the charge is not stored in 2-μF
capacitor. The equivalent capacitance and the charge on the 6-μF and
3-μF capacitors are
Ceq = (1/C1+1/C3)-1 = (1/6+1/3) = 2.00 μF
Q1i = Q3i = Qtot = CeqΔV = 2×10-6×12 = 24.0 μC
The switch 2 is opened before the switch 1 is closed, thus the charge
on the 3-μF does not move. Now 6-μF and 2-μF are connected in
parallel, thus we have
Ceq‘ = C1+C2 = 6+2 = 8.00 μF ΔV’ = Q1i/Ceq’ = 24×10-6/8×10-6 = 3.00 V
Q2f = C2ΔV’ = 2×10-6×3 = 6.00 μC