Download CAPA 2 - Capa Help

1. [1pt]
A person scuffing her feet on a wool rug on a dry day accumulates a net charge of 42.1uC. How many excess electrons does this person get?
Correct, computer gets: 2.63e+14
2. [1pt]
By how much does her mass increase?
Correct, computer gets: 2.40e-16 kg
3. [1pt]
Three charges are arranged on a line as shown above.
(For each statement select T True, F False).
A) If Q1 is positive, Q2 is negative and Q3 is negative, then Q2 MUST feel a net force to the
right.
B) If Q1 is positive, Q2 is negative and Q3 is positive, then Q2 MUST feel a net force to the
right.
C) If Q1 is negative, Q2 is negative and Q3 is positive, then Q2 MUST feel a net force to the
right.
D) If Q1 is positive, Q2 is negative and Q3 is positive, then Q2 MIGHT feel a net force to
the right.
Correct, computer gets: FFTT
4.[1pt]
Two small spheres with charges -4 C and -9 C are held 9.5 m apart. Find the magnitude
of the force between them.
Correct, computer gets: 3.59E+09 N
Hint: This is an application of Coulomb's Law.
5.[1pt]
Four charges Aq,Bq,Cq, and Dq (q = 3.50 x 10-07C) sit in a plane at the corners of a
square whose sides have length d = 63.0 cm, as shown in the diagram below. A charge,
Eq, is placed at the origin at the center of the square.
DATA: A = 4, B = 6, C = 3, D = 6, E = 4. Consider the charge at the center of the square.
What is the x-component of the net force on this charge?
Correct, computer gets: 1.57E-02 N
6.[1pt]
Two point-like particles are placed 26.0 cm apart and are given equal and opposite
charge. The first particle, of mass 12.2 g, has an initial acceleration of 8.85
m/s2 towards the second particle. What is the mass of the second particle if its initial
acceleration towards the first is 8.50 m/s2?
Correct, computer gets: 1.27E+01 g
7. [1pt]
As shown in the figure above, a ball of mass 0.420 grams and positive charge, q =32.8
microC, is suspended on a string of negligible mass in a uniform electric field. We
observe that the ball hangs at an angle of theta =15.0 ofrom the vertical. What is the
magnitude of the electric field?
Correct, computer gets: 3.36e+01 N/C
Hint: Draw a free body diagram for the ball. Since the ball is at rest the vector sum of all
the forces, including the tension in the string, must be zero.
8. [1pt]
The left figure shows seven charges of equal magnitude. Determine the
direction of the total
force on the charge at the origin due to all the other charges. Into which quadrant does
the total force vector point? Use the right diagram to draw the individual
forces (Choose A, B, C or D).
Correct, computer gets: D
9. [1pt]
Two charges, +q and -q, are located in the x-y plane at points (0,+d/2) and (0,-d/2),
respectively. Calculate the magnitude of the electric field at point P with the
superposition principle.
Data: q=38.0 nanoCoulombs,
d=3.80mm and P at
x=76.0mm.
Correct, computer gets: 2.954E+03 N/C
Hint: By the principle of linear superposition, the total electric field at position P is the
vector sum of the electric field contribution from charges +Q and -Q.
10.[1pt]
Two particles of masses 8.7 kg and 5.8 kg and charges 5.9 C and -7.9 C respectively are
separated by a distance 37 cm. Find the ratio of the magnitudes of the Electric to
Gravitational forces between them. That is, what is $F_{elec}/F_{grav}$?
Correct, computer gets: 1.24E+20
Hint: Note that the distance between the particles does not affect the ratio of the forces.
The questions asks for the ratio of the *magnitudes*, so the answer must be positive.
Worked-Out Answers
1. Recall that the charge of an electron is -1.6E-19C or is -1.6E-13µ
# electrons 
ch arg e _ given
ch arg e _ electron
2. Recall that the mass of an electron is 9.11E-31 kg
Total _ mass _ increase # electrons * 9.11 10 31 kg
3. If Q1 is -, Q2 is -, and Q3 is +, then Q2 feels a force to the right from both charges. Therefore, the net force
MUST be to the right.
If Q1 is +, Q2 is -, and Q3 is +, then Q2 will feel a net force towards the smaller charge. Without knowing the
magnitudes of the charge, the net force MIGHT be to the right or left.
4. To solve this problem, Coulomb’s Law is used:
kQ1Q2
r2
k  9.0  10 9
F
5. Be very careful with units in this problem. Keep in mind that Coulomb’s law will be negative since the force
is attractive.
 kQQE
Fq 
r2
2
2
d2 
d  d 
2
r         
2 2
 2 
 2kQQE
d2
For _ ch arg e _ A :
Fq 
FAx   FA cos( 45  ) 
 2kQA QE
* cos( 45  )
d2
For _ ch arg e _ B
FBx   FB cos( 45  ) 
 2kQB QE
* cos( 45  )
2
d
For _ ch arg e _ C
Fcx  Fc cos( 45  ) 
 2kQC QE
* cos( 45  )
2
d
For _ ch arg e _ D
 2kQD QE
* cos( 45  )
d2
2kQ 2 E ( A  B  C  D)
 FAx  FBx  FCx  FDx 
* cos( 45  )
d2
FDx   FD cos( 45  ) 
FNET
6. To solve this problem, use Newton’s third law:
F1  F2
m1 a1  m2 a 2
m2 
m1 a1
a2
7. The units for this problem are in N/C
T
Tcosθ
FE
Tsinθ
mg
Fnet  0
FE  T sin(  )
mg  T cos( )
FE T sin(  )

 tan( )
Fg T cos( )
FE  Eq
Fg  mg
Eq
 tan( )
mg
mg tan( )
E
q
8. Problem 8 varies greatly based on diagram. Remember that like charges repel and unlike charges attract.
They also add as vectors.
9. For E1,
for E2
2
2
d 
r1     x 2
2
E1x  E1 cos( ) 
d 
r2     x 2
2
E1 x
E1x 
E1 y 
E1 d
E1 y   E 2 sin(  )  
E2 d
2 r1
E2 
kq
r1
E2 x
r1
E1 y   E1 sin(  )  
E1 
E1x   E 2 cos( )  
2
E2 
kqx
 d 2

   x2 
 2 



 kqd
3/ 2
 d 2

2    x 2 
 2 



E2 
3/ 2
kq
r2
2
 kqx
 d 2

   x2 
 2 



 kqd
3/ 2
 d 2

2    x 2 
 2 



3/ 2
r2
2 r2
Net and Total E
E xnet  E1x  E 2 x  0
E ynet 
 kqd
 d 2

   x2 
 2 



3/ 2
 Etot
10. For this problem, just use a ratio of the two equations
kQQ
r2
Gmm
Fg  2
r
FE
kqq

Fg Gmm
FE 