Download Exam 1 solutions for PHYS 1215

Phys 1215, First Test. September 20, 2011
50 minutes
Name: ____________________________
Show all work for maximum credit. Each problem is worth 10 points.
k = 9.0 x 109 N m2 / C2
ε0 = 8.85 x 10-12 C2 / N m2
e = 1.602 x 10-19 C
ρ = 1.68 x 10-8 Ω m for copper
1) Calculate the electric force between two alpha particles (charge of 3.2x10-19 C)
separated by a distance of 2.41 x10-10 m.
Fk
19
q 1q 2
C) 2
9
2
2 (3.2  10

9

10
Nm
/C
 16 nN
r2
(2.41  10 10 m) 2
2) Given the two charges shown in the figure, at what position “x” is the electric field
zero? The two charges are separated by a distance of 2 meters.
-4 Q
9Q
2m
For the electric field to be zero: E9Q + E4Q = 0, so E9Q = - E4Q
k
9Q
4Q
9
4
3
2
k 2 
 2 
  3x  4  2 x  x  4 m
2
2
2 x x
( 2  x)
x
( 2  x)
x
3) Find the electric potential at the origin of coordinates due to the charges located at A
and B. Consider the coordinates given in meters and charges QA= 4C and QB= -5C.
QA
4 10 6
 9 10 9
 12,000 volt
r
3
6
QB
9 ( 5  10 )
VB  k
 9  10
 -22,500 volt
r
2
VA  k
So the net electric potential is: VT = VA + VB = 12000 + (-22500) = -10,500 V
4) Determine the electric field at the origin of coordinates due to the charges located at A
and B using the figure in the previous problem. Give your answer as a magnitude and an
angle with the proper units.
First let’s find the magnitudes of the two vectors using the formula for the electric field of
a point charge:
q
Nm 2 4x10 6 C
E A  k A2  9x10 9
 4,000 N/C down since positive charges repel
r
C 2 (3m) 2
EB  k
2
6
qB
C
9 Nm 5x10

9x10
 11,250 N/C to the right since opposite charges attract
2
2
2
r
C
(2m)
Now we consider the directions of these vectors:
EA: Since qA is positive the electric field vector will point away from the charge, so it will
be a vertically downward vector. (0,-4000)
EB: Since qB is negative the electric field vector will point toward the charge, so it will be
a horizontal vector pointing to the right. (11,250,0)
We can add these vectors easily: EA + EB = (11,250, -4000)
E
 Ex    Ey 
2
2
 11,250 2  4,000 2  11,900 N/C
The angle is computed: θ = tan-1 (
EA
EB
) = -20o.
5) A capacitor is made with two plates submerged in water (Kwater=81). If their area is
3.11 m2 and the distance between them is 0.66 mm. Calculate the capacitance (in Farads)
and how much charge is stored in the capacitor when it is connected to a voltage of 120
volts.
C  K o
A
3.11m 2
 (81)(8.85  10 12 F / m)
 3.4 μF
d
0.00066m
And the charge stored is: Q  CV  3.4F(120V)  410 μC
6) Find the power delivered by the voltage source. The voltage is given in RMS
voltage. Also, find the current through resistor R1.
To find the power delivered by the voltage source, notice that the three resistors are
equivalent of one. To find the equivalent, we see that the 12 and the 24 resistors are in
1
parallel, so they are equivalent to one of value Requivalent 
 8 , and then
1
1

12 24
this equivalent resistor is in series with the 10 resistor, giving a total of 18. So the
power is:
(90V ) 2
Power 
 450W
18
To find the current through R1, notice that the total current through the whole circuit is
the same as the current through R1. We don’t know the voltage across R1., but we do
know the voltage across the whole circuit. We can then use Ohm’s Law.
V
90V
I

 5.0 A.
Req
18
7) The switch in the figure has been closed for a long time, so the capacitor is charged to
30 volts. Then at time t = 0 you open the switch and the voltage starts dropping (the
multimeter behaves as a resistance of 1.5MΩ) find how long it takes for the voltage in the
capacitor to get to 5V.
The time constant is   RC  (1.5M)(18F )  27seconds
The equation for discharge is: Vc  Vet / 
With the values of the problem: 5  30e t / 470
5
 5
 5
Solving for t we get:
 e t / 27  ln    t / 27  t  27 ln    48 seconds
30
 30 
 30 
8.) Determine the capacitance of the following arrangement of capacitors and find how
much energy is stored when you apply 24V between terminals A and B.
To determine the capacitance of this arrangement of capacitors notice that the branch on
the left side is made of 3 capacitors in series. That branch is equivalent to one capacitor
of value:
1
C
 6F
1
1
1


18F 18F 18F
Then, this equivalent capacitor is in parallel with an 18µF capacitor giving a total of
24µF together.
And the energy stored is: E 
1
1
CV 2  24nF(24V) 2  6.9 mJ
2
2
9) Capacitance, Dielectrics and Storage of Electric Energy.
A cardiac defibrillator is used to shock a heart that is beating erratically. A capacitor in
this device is charged to 3.5 kV and stores 1300 J of energy. What is the capacitance?
1
CV 2
2
the capacitance is:
Energy 
C
2  Energy 2  1300J

 212 µF
V2
(3500V) 2
10) We want to find the current through the resistors R1, R2 and R3.
Draw the direction of the currents on the diagram.
Write down the equations that you need to solve the problem.
You will get full credit if the equations are correct (signs, quantities, etc.)
You do not need to solve the equations.
-
+
I1
-
+
I3
+
I2
-
+
The node equation is:
I2 = I1 + I3
The loop equations are:
10  35I1  45I 2  0
 5  5I 3  45I 2  15I 3  0
left loop (I chose clockwise)
right loop (I chose clockwise)
And writing them in the standard form:
- I1 + I2
- I3 = 0
35I1  45I 2 ..........  10
45I 2  20I 3  5