Download ANSWERS TO THE HOMEWORK FROM THE BOOK FOR THE

ANSWERS TO THE HOMEWORK FROM THE BOOK FOR THE FIRST DAY’S ASSIGNMENT
FOR ELECTRICITY.
 Calculus: Skim/Read pp 601 to bottom header 613 AND bottom header (19.6) 616 to bottom
header (19.8) 621HW: Problems #4, 5, 11, 13, & 15 starting page 635
4. F= kqq/r2 = 9E9(1.6 E –19)2-/2E –152 = 57.6 N
5. F of the – 4 C charge = 9E9(4E-6)(7E-6)/0.52 = 1.008 N 60 S of E
F of the +2 C charge = 9E9(2E-6)(7E-6)/0.52 = 0.504 N 60 N of E
Using addition of vectors by components
Rx= +0.252 + (+0.504) = 0.756 N (East)
Ry = -0.8730 + (+0.4365) = -0.4365 N (South)
R = 0.7562  0.43652 = 0.873 N
 = tan-1(.4365/756) = 30
ANSWER: 0.873 N 30 South of East
11.
------------------------
*------ -2.5 ---------------6 C
( x
)(
1m
)
WE HAVE GOT TO BE ABLE TO FIGURE OUT DIRECTION OF ELECTRIC FIELD. This is a
very, very important concept. There are two ways. (1) Remember that E-field lines always point
OUT OF A POSITIVE CHARGE AND INTO A NEGATIVE CHARGE. (2) Remember the
direction of the electric field is decided as if you had placed a small POSITIVE test charge wherever
you are determining the electric field. Ask yourself which way it will move because of the other
charge and that is the direction of the electric force AND E force and E field always point the same
way for a positive charge.
The only place that the electric field can be zero due to the two charges is on the left of the –2.5C
BECAUSE At any point in between the electric field due to the –2.5 C points left and the electric
field due to the 6C also points left so they can’t cancel each other out. REMEMBER electric field is
a vector! On the right of the 6 C charge the electric field due to the 6C points right and the E field
due to the –2.5 C points left but the E field due to the 6 WILL ALWAYS BE BIGGER because it is
always closer (E=kq/r2). On the left the E field due to the 2.5 will point right and the E due to the 6
will point left and at some point they will equal each other since the 6 is bigger, but further away and
the 2.5 is smaller, but closer. So:
Let x equal the distance between the –2.5 and the place where the E field is zero.
Therefore 1+x will be the distance b between the 6 and the place where the E field is zero.
The electric field due to the 6 and the one due to the 2.5 will be equal and opposite: Mathematically:
9 E 9(2.5 E  6) 9 E 9(6 E  6)
=
Divide both sides by 9E9 and 1E-6 then cross multiply
x2
(1  x) 2
2.5(1+x)2 = 6x2
Expand and put everything on one side
2.5 + 5x + 2.5x2 = 6x2
3.5 x2 - 5x – 2.5 = 0
Dang it I wish I had that program, but I will use the quadratic
formula instead, a = 3.5, b = -5, c = -2.5
ANSWER: x = 1.82 or –0.391 m. The negative root will put the point between the two charges so it
the answer must be 1.82 meters to the left of the 2.5 c charge.
13. E field due to the 6 at the origin = (9E9)(6E-9)/0.302 = 600 N/C west
E field due to the 3 at the origin = (9E9)(3E-9)/0.12 = 2700 N/C south
Using component addition E = 600 2  2700 2 = 2766
 = tan-1 (2700/600) =77.5
(a) ANSWER = 2766 N/C 77.5 South of West
(b) E = F/q; F = Eq = 2766 (5E-9) = 1.38 E –5 N 77.5 South of West (same directions because it
is a positive charge
Darn ijk vector notation and who uses kilonewtons and micronewtons in a problem where charge
is given in nanocoloumbs! My answers are actually the same as the book’s!
E of 2q = 2kq/a2 east
E of 3q = 3kq/(2a2) 45 N of E
E of 4q = 4kq/a2 north
Using addition of vectors by components
Rx = +2kq/a2 + (3/2) kq/a2 cos (45) = 3.06 kq/a2
Ry = +4kq/a2 + (3/2) kq/a2 sin(45) = 5.06 kq/a2
ANSWER TO (a) That gives ijk vector notation (Rx is i and Ry is j), but just in case you want
numbers you are used to: R = 5.91 kq/a2 58.8 N of E
Answer to b is to just multiply all answers to (a) by q!
15.

Algebra: Skim/Read pp 497 to 517. Only Skim/Read pp 517 to 522 for entertainment or to get a
taste of what the calculus kids will have to do pretty soon! Ha ha! HW: Problems # 4, 8, 11, 24,
27, starting on page 525
4. F due to charge in NW corner = kq2/a2 North
F due to charge in NE corner = kq2/2a2 45 North of East
F due to charge in SE corner = kq2/a2 East
Using components
Rx = +kq2/2a2 cos(45) + (+kq2/a2) = 1.354 kq2/a2
Ry = +kq2/2a2 sin(45) + (+kq2/a2) = 1.354 kq2/a2
R=
R x2  R y2 = 1.91 kq2/a2 and the angle is 45 N of E since both Rx and Ry are the same.
8. Force of gravity = 9.11 E –31 (9.8) = Electric force = (9E9)(1.6E-19)2/r2
solving for r, r= 5.08 m
11. Find the magnitude of the force the 3nC exerts on the 5 using Coulomb’s law. The direction is
south (why?). Find the magnitude of the force the 6nC exerts on the 5 using Coulomb’s law. The
direction is West (why?). Then add those two vector forces using algebraic addition of vectors by
components. The answer is 1.38 E –5 N 77.5 South of West.
24. Okay Fun times.
E field due to the 8nC charge = (9E9)(8E-9)/0.252 = 1152 N/C East
E field due to –5nC charge = (9E9)(5E-9)/0.252 = 720 N/C East
Before we can find out the E field due to the 3nC we have to figure out how far it is away from the
point on the x axis midway between the two lower charges. The distance is one leg of a right triangle
so it = 0.5 2  0.25 2 = 0.433 m SO:
E field due to the 3nC charge = (9E9)(3E-9)/0.4332 = 144 N/C South
Using addition of vectors by components
Rx= +1152 + (+720) = 1872 (East)
Ry = 144 (South)
R = 1872 2  144 2 = 1878 N/C  = tan-1(144/1878) = 4.38
ANSWER: 1878 N/C 4.38 South of East
27. This problem is the exact same as #11 for the calculus students. See the answer shown above for
#11 for them.