Homework Assignment #2 - Thursday June 16, 2005
1. (a) On a piece of graph paper (remember that some are available on the course web
page) place a  3 .C charge at the origin and a  2.C at the point ÐB œ 3.0 mß C œ 0Ñ.
(b) Determine the magnitude of the electric forces acting on each of the two charges and
sketch these forces on your diagram. Write down your force magnitudes below:
Magnitude of force on  3 .C charge:
J œ 5l ;1 ;2 lÎ<2 œ Ð9 ‚ 109 N m2 C2 ÑÐ3 ‚ 106 CÑÐ2 ‚ 106 CÑÎÐ3.0 mÑ2 œ 6.0 ‚ 103 N
Magnitude of force on  2 .C charge:
By Newton's third law, this must be the same: 6.0 ‚ 103 N
2. Repeat for the following situations. You can use the same piece of graph paper if you
plan out your scale properly and place the origin at a new position
(a) a  3 .C charge at Ð0ß 2.0 mÑ and a  2 .C at the point ÐB œ 3.0 mß C œ 2.0 mÑ
Magnitude of force on  3 .C charge:
J œ 5l ;1 ;2 lÎ<2 œ Ð9 ‚ 109 N m2 C2 ÑÐ3 ‚ 106 CÑÐ2 ‚ 106 CÑÎÐ3.0 mÑ2 œ 6.0 ‚ 103 N
Magnitude of force on  2 .C charge:
By Newton's third law, this must be the same: 6.0 ‚ 103 N
(b) a  4 .C charge at Ð0ß  2.0 mÑ and a  2 .C at the point Ð  3.0 mß  2.0 mÑ
Magnitude of force on  2 .C charge:
J œ 5l ;1 ;2 lÎ<2 œ Ð9 ‚ 109 N m2 C2 ÑÐ4 ‚ 106 CÑÐ2 ‚ 106 CÑÎÐ3.0 mÑ2 œ 8.0 ‚ 103 N
Magnitude of force on  4 .C charge:
By Newton's third law, this must be the same: 8.0 ‚ 103 N
3. On your old graph paper or a new one, place a  3 .C charge at the origin and a
charge of  2 .C charge at the point ÐB œ 3.0 mß C œ 4.0 mÑ.
(a) Sketch the directions of the forces on each charge and determine their magnitudes.
Magnitude of force on  3 .C charge:
The distance between the two charges is ÈÐ3.0 mÑ2  Ð4.0 mÑ2 œ 5.0 m
J œ 5l ;1 ;2 lÎ<2 œ Ð9 ‚ 109 N m2 C2 ÑÐ3 ‚ 106 CÑÐ2 ‚ 106 CÑÎÐ5.0 mÑ2 œ 2.2 ‚ 103 N
Magnitude of force on  3 .C charge:
By Newton's third law, this must be the same: 2.2 ‚ 103 N
(b) Determine the B and C components of the forces on each charge. You can determine
the necessary sines and cosines from the components of the displacement between the
charges.
 3 .C charge has force:
B component:
First, note that for the 3-4-5 right triangle formed by the force and sides parallel to the B
and C axes, cos ) œ 3/5 and sin ) œ 4/5.
JB œ J cos ) œ  Ð2.16 ‚ 103 NÑÐ3Î5Ñ œ  1.3 ‚ 103 N
C component:
JC œ J sin ) œ  Ð2.16 ‚ 103 NÑÐ4Î5Ñ œ  1.7 ‚ 103 N
 2 .C charge has force:
B component:
JB œ  1.3 ‚ 103 N (the negative of JB for the  3 .C charge)
C component:
JC œ  1.7 ‚ 103 N (the negative of JB for the  3 .C charge)
4. On your old graph paper or a new one, place the following charges:
A  3 .C charge at the origin
A  2 .C charge at (3.0 m, 0)
A  3 .C charge at (0, 3.0 m)
Determine the B and C components of the two electric forces acting on the charge at the
origin, and then determine the total electric force acting on it, giving it first in terms of its
B and C components and then in terms of magnitude and direction. Write your answers
below:
B component of total force:
The only force with an B component is the attractive force between the  3 .C charge at
the origin and the  2 .C charge on the positive B axis; thus
JB œ 5l;1 ;2 lÎ<2 œ Ð9 ‚ 109 N m2 C2 ÑÐ3 ‚ 106 CÑÐ2 ‚ 106 CÑÎÐ3.0 mÑ2
œ  6.0 ‚ 103 N
C component of total force:
The only force with a C component is the repulsive force between the  3 .C charge at
the origin and the  3 .C charge on the positive C axis; thus
JC œ 5l;1 ;2 lÎ<2 œ  Ð9 ‚ 109 N m2 C2 ÑÐ3 ‚ 106 CÑÐ3 ‚ 106 CÑÎÐ3.0 mÑ2
œ  9.0 ‚ 103 N
Magnitude of total force:
J œ ÈÐ  6.0 ‚ 103 NÑ2  Ð  9.0 ‚ 103 NÑ2 œ 10.8 ‚ 103 N
Direction of total force:
The force is at an angle of ) œ tan1 Ð  9.0Î6.0Ñ œ tan1 Ð  1.5Ñ œ  56° (equivalent
to  304°), close to a southeastern direction when north is the  C direction and east is the  B
direction.
Sketch the electric force contributions and the total electric force on your diagram, and
check that the results are consistent with your calculations.