---~---ãããã--ããã-~----
r
then the electric force between them, FE, is directed along the line joining them. The magnitude of this force is proportional to the charges (q1 and q2) and inversely proportional to r2" as given by
COULOMB'S LAW FE =k q1q2
r2
The proportionality constant is k, and in general, its value depends on the material between the particles. However, in the usual case where the particles are separated by empty space (or by air, for all practical purposes), the proportionality constant is denoted by k0 and called Coulomb's constant. This is a fundamental constant of nature (equa,l in magnitude, by definition, to 10-7 times the speed of light squared), and its value is 9 x 109 Nãm2 /C2:
Coulomb's constant
ko = 9 x 109 Nãm2/C2
This is the value of k you will always use (unless you're specifically given another value, which would happen only if the charges were embedded in some insulating material that weakens the electric force).
If we retain the signs of the charges q1 and q2 when we use the formula FE= kq1q2/r2, then a positive FE means that the particles repel each other; a negative FE means they attract each other.
This is consistent with the fact that like charges (two positives or two negatives) repel each other, and opposite charges (one positive and one negative) attract. Note that the two electric forces in each of the following diagrams form an action/ reaction pair.
148 MCAT PHYSICAL SCIENCES REVIEW
. ~FE on+ 8--- 0 FE on+.
~FE on-0---________________ ----_-(:)FE on-.
OFEon+ FEon-f:""\
--- • --- • \::..J---
---OFEon-• --- • FEon+G + 'ãã-ã--ã
like charges repel
opposite charges attract
~ Example 6-3: Two charges, q1 = -2 x 10-6 C and q2 = +5 x 10-6 C, are separated by a distance of 10 em. Describe the electric force between these particles.
Solution. Using Coulomb's Law, we find that
FE = ko ql ~2 = (9 X 109 Nã~2) ( -2 X 10-6 C)( +5 >< 10-6 C) -
r c (lo-t m)2 --9 N
The minus sign on FE tells us that the electric force between the charges is attractive, which we expected since q1 is negative and q2 is positive. Therefore, the electric force on q1 is directed toward q2, the electric force on q2 is directed toward q1, and the magnitude of the electric force that each charge feels due to the other is 9 N.
q 1 0
\.J FEonq1
Gq2
Eonq2 \.J
~ Example 6-4: A coulomb is a lot of charge. To get some idea just how much, imagine that we had two objects, each with a charge of 1 C, separated by a distance of 1 m. What would be the electric force between them?
Solution. Using Coulomb's Law, we'd find that
p, = z.. qtq2 =(9x109 Nãm2) (1C)(1 C) =9x109 N E ~ r2 . c2 (1 m)2
To write this answer in terms of a more familiar unit, let's use the fact that 1 pound (1 lb) is about 4.5 N, and 1 ton is 2000 lb:
( .
1 9 N) 1lb 1 ton . 1.
1. FE= 9x 0 ã - - ã =one rm ton tons
4.5 N 2000 lb
MCAT PHYSICS - CHAPTER 6: ELECTROSTATICS 149
That's equivalent to the weight of about twenty-five hundred Boeing 747s! It's now easy to understand why most real-life situations deal with charges that are very tiny fractions of a coulomb; the microcoulomb (1 f,LC = 10-6 C) and the nanocoulomb (1 nC = 10-9 C) are more common "practical" units of charge .
.... Example 6-5: Consider a charge, +q, initially at rest near another charge, -Q.
How would the magnitude of the electric force on +q change if -Q were moved away, doubling its distance from+q?
Solution. Coulombis Law is an inverse-square law, FE ex 1/r2, so if r increases by a factor of 2, then FE will decrease by a factor of 4 (because 22 = 4) .
.... Example 6-6: Consider two plastic spheres, 1 meter apart: a little sphere with a mass of 1 kg and an electric charge of + 1 nC, and a big sphere with a mass of 11 kg and an electric charge of +11 f,LC.
Solution.
(a) Find the electric force and the gravitational force between these spheres. Which force is stronger?
(b) If the big sphere is fixed in position, and the little sphere is free to move, describe the resulting motion of the little sphere if it's released from rest.
(a) Using Coulomb's Law, we find that the electric force between the spheres is FE = ko q~ = (9 x 109 N~~2) (1 x 10-9 C)(11 .. x 10-6 C) = 9.9 x 10-s N == 10-4 N
r (1m)
Using Newton's Law of Gravitation, the gravitational force between them is E =GmM =(6.7x1o-n Nãm2) (1 kg)(11 kg) ==7.4x1o-Io N
G r2 kg2 (1 m)2
Which force is stronger? It's no contest: The electric force is much stronger than the gravitational force. So, even though the spheres experience an attraction due to gravity, it is many orders of magnitude weaker than their electrical repulsion and can therefore be ignored.
(b) The net force on the little sphere is essentially equal to the electrical repulsion it feels from the big sphere (since the gravitational force is so much smaller, it can be ignored). Therefore, the initial acceleration of the little sphere is
FE-10-4 N ~10-4 m/s2
a=-;;- 1kg
directed away from the big sphere. Notice that as the little sphere moves away, its acceleration does not remain constant. Because the electric force is inversely proportional to the square of the distance between the charges, as the little sphere moves away, the repulsive force it feels weakens, so its acceleration decreases.
Therefore, the little sphere moves directly away from the big sphere with decreasing acceleration.
150 MCAT PHYSICAL SCIENCES REVIEW
0
8 0 q As the little sphere moves away, ..
the force it feels and its acceleration both decrease.
Nevertheless, because the acceleration of the little sphere always points in the same direction (namely, away from the big sphere), the speed of the little sphere is always increasing, although the rate of increase of speed gets smaller as the little sphere gets farther away.
THE PRINCIPLE OF SUPERPOSITION FOR ELECTRIC FORCES
"The deepest principle in electrodynamics ... is the superposition principle. n
- Richard Feynman
Coulomb's Law tells us how to calculate the force that one charge exerts on another one. But what if two (or more) charges affect a third one? For example, what is the electric force .on q3 in the following figure?
q10
r
q30 8q2
R
Here's the answer: If Ft-on-3 is the force that q1 alone exerts on q3 (ignoring the presence of q2) and if F2_on-3 is the force that q2 alone exerts on q3 (ignoring the presence of q1), then the total force that q3 feels is simply the vector sum Ft-on-3 + F2_0n_3• The fact that we can calculate the effect of several charges by considering them individually and then just adding the resulting forces is known as the principle of superposition. (This important property will also he used when we study electric field vectors, electric potential, magnetic fields, and magnetic forces.)
The Principle of Superposition The net electric force on a charge ( q) due to a collection of other charges (Q's)
is equal to
th~ sum of the individual forces that each of the as alone exerts on q.
MCAT PHYSICS- CHAPTER 6: ELECTROSTATICS 151
_q10
r
q30 eq2
R
~ Example 6-7: In the figure above, assume that q1 = 2 C, q2 = -8 C, and q3 = 1 nC.
If r = 1 m and R = 2 m, which one of the following vectors best illustrates the direction of the net electric force on q3?
B./ A.'
c. "
D./
Solution. The individual forces F1_
0n_3 and F2_0n_3 are shown in the figure below.
Adding these vectors gives F0n3, which points down to the right, so the answer is C.
F, = k ql q3 = (9 X 109 N;m2) (2 C)(1 X 10-9 C)
1-on-3 o ,2 c2 (1 m)2
q10
= 18 N (repulsive; away from q1)
F2-on-3 F2-on-3 = ko q2;3 = (9 X 109 Nã~2) ( -8 C)(1 X 10-9 C)
R c (2 m)2
= -18 N (attractive; toward q2)
If the question had asked for the magnitude of the net electric force on q3, then we'd use the Pythagorean theorem to find the length of the vector F on 3. The vector F on 3 is the hypotenuse of the right triangle whose legs are F1-on-3 and F2_0n_3, so the
magnitude of F on 3 is found like this:
(Fon3)2
= (fi-on-3)2
+ (F2-on-3)2
= 182 +182
= (182)(2) :. Fon3 = 18..J2 N
:::25N
F2-on-3
A '
~~-~-w }
+ -- I
q3 I
, ( ' 0.., I I 18 N F1-on-3
Ul:!l. "'\~ :
o>~,
"V
Gq2
152 MCAT PHYSICAL SCIENCES REVIEW
~Example 6-8: In the figure below, assume that q1 = 1 C, q2 = -1 nC, and q3 = 8 C.
If q4 is a negative charge, what must its value be in order for the net eleCtric force on q2 to be zero?
~ ~ ~ ~
0---ã---8---0---8
1m 2m 1m
Solution. The individual forces F1-on-2, F3_0n_2, and F4-on-2 are shown in the figure below. Notice that ft-on-2 and F4-on-2 point to the left, while F3_on_2 points to the right.
Q3 q4
q1 F1-on-2 q2 F
f.::\ . . 3-on-2
\:)---.. ---8 . .. ---(!)---<:)
F4-on-2
If we let q4 = ...:.x C, then the magnitudes of the individual forces on q2 are l&-on-2l=k
0 q1jq21 =(9x109 Nã~1) (1 C)(1 nC) _ (rl-2)2 cz (1 m)2 -9 N JF3-on-21 = ko . q2q3 = (9 x109 Nãm1) (1l).C)(8 C)
(r2-3)2 c:z (2m)2 =18N IF4-on-21ã=ko q 2 Jq~l =(9x109 Nãm1) (1 nC)(xC)
(r2-4 )2 c:z (3 m)2 = x N
In order for the net electric force on q2 to be zero, the sum of the magnitudes of FI-on-2 and F 4-on-2 must be equal to the magnitude of F3-on~2• That is,
9N +xN= 18N sox= 9. Therefore, q4 = -x C = -9 C.
MCAT PHYSICS - CHAPTER 6: ELECTROSTATICS 153