Download Electrostatics Review Sheet

Electrostatics Review Sheet
AP Physics
Conceptual Review
1. How are charged objects fundamentally different from uncharged objects?
The difference between charged objects and uncharged objects is that charged objects are capable
of exerting an electric force while uncharged objects are not.
2. If all objects are composed of charged object (electrons and protons), why are we usually not aware
of this fact?
We usually do not notice that we are made up of charged objects because there are two types of
charge – positive and negative – and most objects in the world are made up of an equal number of
positive and negative charges. This means that when the charged particles push and pull on each
other, they pull on one type of charge and push equally hard on the other type. The net effect is that
the pushes and pulls cancel out, leaving it as if there were no force.
Interestingly, most forces between two objects occur because of the charged particles in one body
interacting with charged particles in another body. When the objects get very close, the electrons in
the outer layers of each object’s electrons are the parts that get closest. Since the electric force gets
stronger as the charges get closer, then the repulsion between the electrons becomes stronger than
the attraction toward protons and the net force is to stop the objects from getting any closer. This is
what we call touching.
3. How do you pick up a negative charge by scuffing your sneakers on the carpet? What is happening
while you drag your feet across the floor?
This is an example of charging by friction. As your shoes rub across the carpet, the rubber pulls
electrons off of the carpet material. This means that your shoes and your body end up with extra
electrons compared to protons and an overall negative charge. Of course, the carpet ends up with
fewer electrons than protons and an overall positive charge.
4. You pick up a positively charge metal ball that is resting on an insulated surface and get shocked
when you touch it. What happens to create the shock? What is different about you after the shock?
This is an example of conduction. The shock happens as electrons jump from your body because
of they are more attracted to the positively charged ball. As the electrons leave your body, you end
up with fewer electrons than protons, which means that you have an overall positive charge. The
ball was positively charged because it had fewer electrons than protons, but as it steals your
electrons it ends with less overall charge than it started with.
5. While you are hiking in the mountains, a thundercloud with extra negative charges on its lower level
forms close to you. You feel your arm hairs start to stand up and run for cover. While ducking under a
ledge, you grab a metal handrail embedded in the ground. How are you charged after grabbing the
rail? Why?
This is an example of induction. Your hair is standing up because you are getting polarized by
the negative thundercloud overhead. The electrons in your body move away from the thundercloud,
down into your feet, leaving you with fewer electrons than protons – and a positive charge – in the
upper part of your body. When you grab the handrail, you ground yourself and allow your electrons
to move into the handrail and the earth – getting even further away from the negative cloud. After
you let go of the handrail, you can’t get your electrons back. You now have fewer electrons than
protons in your body, so you have a net positive charge.
6. How does the electric force compare to the gravitational force? How are they alike? How are they
different?
The electric force is similar to the gravitational force in in that they both are capable of attracting
two object together and they both obey very similar equations.
The electric force is different from the gravitational force because it works between charged
objects while the gravitational force works between massive objects. This leads to another
difference because there are two types of charge and only one type of mass. The two different types
of charge can create forces that repel as well as attracting, while gravity can only attract. In addition,
the electric force is fundamentally a lot stronger than the gravitational force – you can begin to see
why if you notice how much bigger the constant in Coulomb’s Law is than the constant in Newton’s
Law of Universal Gravitation.
7. What is an electric field? Why was the idea of an electric field created?
The idea of an electric field was created to explain how two charged particles could exert forces on
each other without ever touching. The electric field is the way that one charge particle reaches out
and affects another charged particle. You can think of it as extending around a charged particle like
the tentacles of an octopus extend around it’s body, except that there are more than eight tentacles
and they reach out to every point in space
8. Draw a sketch of the field lines around a:
a) positive charge
b) negative charge
c) nearby positive and negative charge
9. What is special about the electric field between two charged parallel plates?
The electric field between two charged parallel plates is constant – it has the same strength and
direction everywhere between the plates (as long as you stay away from the very edges). This is
very distinctive from the field around a point charge, which points in different direction on different
sides and gets weaker as you get further away.
10. Why is the static electric field inside a conductor always equal to zero? What would happen if there
ever were an electric field inside a conductor?
A conductor is a material where charges (almost always electrons) move freely. Whenever there
is a field inside a conducting material, it exerts a force on the electrons and moves them to one side
of the object – polarizing the object. The separated charges inside the polarized object create their
own internal field, which point in the opposite direction to the external field. Electrons will keep
moving until their field is strong enough exactly cancels out the external field, leaving a total field of
zero inside the object.
Calculations
1. While driving, your car picks up a net charge of –4 C. How many extra electrons does it have?
Since every electron carries a charge of 1.6 x 10-19 C, we just need to divide the total charge to
-4 x 10-6 C
figure out how many pieces it takes to make it up: -1.6 x 10-19 C = 2.5 x 1013 extra electrons
This may seem like a lot of electrons, until you realize that there something like 3 x 1029 electrons
inside your car, so the extra electrons are just a drop in the bucket.
2. The proton and electron in a hydrogen atom are usually separated by a distance of 5.3 x 10-11 m.
a) How strong is the electric force between the two particles?
Fe =
kq1q2 (9 x 109 Nm2/C2)(1.6 x 10-19 C)(-1.6 x 10-19 C)
=
= –8.2 x 10–8 N
r2
(5.3 x 10-11 m)2
b) Calculate the acceleration of each particle under that force.
F
8.2 x 10–8 N
electron: a = m = 9.11 x 10-31 kg = 9.0 x 1022 m/s/s
F
8.2 x 10–8 N
proton: a = m = 1.67 × 10-27 kg = 4.91 x 1019 m/s/s
c) Challenge – If the electron orbits around the proton in a circle, how quickly does the electron have to
be moving to stay in an orbit that size?
For the electron to be in orbit around the proton, there needs to be a centripetal force keeping it
turning in a circle. The electron’s attraction to the proton is that centripetal force.
mv2
Fe = FC = r which we rearrange to get:
v=
r·FC
m=
5.3 x 10-11 m·8.2 x 10–8 N
9.11 x 10-31 kg = 2.18 x 106 m/s
3. The force between two identical charges, q, is F when they are a distance, d, apart. How big is the
force is one of the charges is doubled and they are moved three times as far apart?
kq1q2
r2
kq2
In the case we are given F = d2
k(2q)q 2kq2 2 kq2 2
For the new case: Fe = (3d)2 = 9d2 = 9· d2 = 9 F so the force is two ninths as large as before.
Coulomb’s Law says: Fe =
4. A hamster in a hamster ball that is charged to 75 C is
released in a room with charged obstacles. At one point,
the hamster is 90 cm south of a pillar that is charged to 400
C and 120 cm east of a pillar that is charged with 700 C.
What is the net force on the hamster ball?
The situation is shown in the picture to the right. Each
of the pillars will exert an electrical force on the ball:
kq1q2 (9 x 109 Nm2/C2)(75 x 10-6 C)(700 x 10-6 C)
F1 = r2 =
= 328 N
(1.2 m)2
kq1q2 (9 x 109 Nm2/C2)( 75 x 10-6 C)(400 x 10-6 C)
F2 = r2 =
= 333.3 N
(0.9 m)2
Since both forces are positive, they are both repulsive forces. A free body diagram for the hamster
ball would look like this:
To find the total force on the hamster ball, we have to add up all of the forces (as vectors):
(Fnet)2 = (328 N)2 + (333.3 N)2
(Fnet)2 = 218672.89 N2
Fnet = 467.6 N
333.3 N
tan  = 328 N
333.3 N
 = tan-1 328 N  = 45.5° S of E


So the total force is 467.6 N at 45.5° S of E
5. In an electronic circuit, three ball that are mounted 2 cm apart in an
equilateral triangle (see picture), acquire a charge as the circuit operates.
The northern ball builds up a charge of +0.2 C. The eastern ball builds up a
charge of –0.7 C and the western ball builds up a charge of +0.5 C. What
is the net electric force on the western ball?
The western ball will be affected by the electric forces from the other two
balls:
kq1q2 (9 x 109 Nm2/C2)(0.5 x 10-6 C)(-0.7 x 10-6 C)
F1 = r2 =
= -7.875 N
(0.02 m)2
kq1q2 (9 x 109 Nm2/C2)(0.5 x 10-6 C)(0.2 x 10-6 C)
F2 = r2 =
= 2.25 N
(0.02 m)2
F1 is attractive and F2 is repulsive, giving us a free body diagram like so:
We know that the angled force is at a 60° angle because it repels
along one side of an equilateral triangle. Every angle inside an
equilateral triangle is 60°, so any force going along that line is at
the same angle. In this specific case, the angle is 60° S of W
To find the total force, we must break the angled force into components, add up the components and
then add up the net components.
Horizontal component:
x = 2.25 N cos60 = 1.125 N to the west
Vertical component:
y = 2.25 N sin60 = 1.95 N to the south
Total Horizontal Force
1.125 N, W + 7.875 N, E = 6.75 N, E
Total Vertical Force
1.95 N, S = 1.95 N, S
Add the net components to find the total force:
(Fnet)2 = (6.75 N)2 + (1.95 N)2
(Fnet)2 = 49.365 N2
Fnet = 7.03 N
1.95 N
tan  = 6.75 N
1.95 N
 = tan-16.75 N = 16.11° S of E


The net force is 7.03 N at 16.11° S of E
6. How strong is the electric field created by a +40 mC charged ball, if you are:
A charged ball will act like a point charge from the outside, so we can use the point charge formula.
a) 40 cm away?
kq (9 x 109 Nm2/C2)(40 x 10-3 C)
E = r2 =
= 2.25 x 109 N/C
(0.4 m)2
b) 100 cm away?
kq (9 x 109 Nm2/C2)(40 x 10-3 C)
E = r2 =
= 3.6 x 108 N/C
(1 m)2
c) 5 m away?
kq (9 x 109 Nm2/C2)(40 x 10-3 C)
E = r2 =
= 1.44 x 107 N/C
(5 m)2
7. If objects with the following amounts of charge is placed 5 m away from the ball in problem 6, how
strong a force will each one experience?
In problem 6, we calculated the strength of the electric field 5 m away from the ball. Let’s use those
results to make our lives simple now. You could use the Coulomb Law if you were a masochist.
a) 4 C
Fe = E·q = 1.44 x 107 N/C · 4 x 10-6 C = 57.6 N
b) 20 C
Fe = E·q = 1.44 x 107 N/C · 20 x 10-6 C = 288 N
c) 300 C
Fe = E·q = 1.44 x 107 N/C · 300 x 10-6 C = 4320 N
8. Point P is located 2 m south of one charged ball and 1.7 m east of another
charged ball. The net field at point P is 149613 N/C at 56° S of W. What is the
charge on each of the balls?
Two unknown charges are creating an electric field at point P. The net field
pulls at an angle, though the individual fields point in cardinal directions.
To figure out the field created by each charge, we need to separate the net
field into components and figure out which component goes with each charge:
Horizontal component
x = 149613 N/C · cos(56) = 83700 N/C west (toward q2). This
component of the field is produced by q2 – which must be negative since
the field is pointing toward it.
kq
E·r2 83700 N/C·(1.7m)2
E = r2 → q2 = k = 9 x 109 Nm2/C2 = –2.67 x 10-5 C
Vertical component
y = 149613 N/C · sin(56) = 124,000 N/C south (away from q1). This
component of the field is produced by q1 – which must be postive since
the field points away from it.
q=
E·r2 124000 N/C·(2 m)2
-5
k = 9 x 109 Nm2/C2 = 5.51 x 10 C
9. A plasma cannon defending a colony on Mars is powered by two electrostatic generators, which are
built 30 apart along an east-west line. While the cannon is operating, the generators build up some
wicked charge – the eastern one reaches -40 mC and the western one builds up -55 mC.
a) What is the total electric field at a point 20 meters above the eastern generator?
The total electric field is the (vector) sum of the
fields created by each individual generator.
The field created by the eastern generator is:
kq 9 x 109 Nm2/C2 · -40 x 10-3 C
E = r2 =
(20m)2
5
= -9.0 x 10 N/C, pointing downward
toward the negative charge
The western generator is similar, but I need to
find its distance from the point first:
c2 = (30m)2 + (20 m)2 = 1300 m2
c = 36.1 m
kq 9 x 109 Nm2/C2 · -55 x 10-3 C
E = r2 =
= -3.8 x 105 N/C, pointing diagonally toward the negative
(36.1 m)2
charge. The angle will be the same as the angle in the triangle drawn above:
30
 = tan-120 = 56.3° west of down
 
To total up the two fields, I must first break the angled field into components:
Horizontal Component:
x = 380,000 N/C · sin(56.3) = 316,000 N/C, West
Vertical Component
y = 380,000 N/C · cos(56.3) = 211,000 N/C, Down
Horizontal
316,000 N/C, West
Then find the net field along each component:
Vertical
211,000 N/C, Down + 900,000 N/C, Down
= 1,111, 000 N/C, Down
Then add up the net components to find the total field (picture to right):
(Enet)2 = (1,111,000 N/C)2 +(316,000 N/C)2 = 1.334 x 1012 N2/C2
Enet = 1.155 x 106 N/C
1.111 x 106 N/C
 = tan-1 3.16 x 10 5 N/C  = 74.1° below west


The net field is 1.155 x 106 N/C at 74.1° below west
b) If a plane carrying 2 mC of charge passes 20 meter above the eastern
generator, what will the electric force on it be?
An electric field pulls a positive charge in the same direction as the field:
Fe = E · q = 1.155 x 106 N/C · 2 x 10-3 C = 2310 N at 74.1° below west
10. A pair of circular plates, 3 m in diameter, are suspended 20 cm apart, parallel to each other. Then
3.1 x 1016 electrons are then moved from the bottom plate to the top plate.
a) How strong is the electric field between the two plates?
The unbalanced charge is 3.1 x 1016 electrons x 1.6 x 10-19 C/electron = 4.96 mC
The formula for the electric field between two parallel plates is:
4kQ 4·9 x 109 Nm2/C2 · 4.96 x 10-3 C
E= A =
= 7.94 x 107 N/C
(1.5m)2
b) How much force would an electron experience of it was placed halfway between the plates?
We can find the force by looking at the effect of the plates’ field on the electron:
Fe = E·q = 7.94 x 107 N/C · 1.6 x 10-19 C = 1.27 x 10-11 N
c) How much force would an electron experience if it was placed 1 cm above the bottom plate?
Because the electric field between the plates is constant, the electrons should experience the same
force anywhere between the plates: 1.27 x 10-11 N
d) How much force would an electron experience if it was placed 1 cm below the top plate?
Because the electric field between the plates is constant, the electrons should experience the same
force anywhere between the plates: 1.27 x 10-11 N
11. A pair of square parallel plates, 1 m on a side are charged to +36 C and –36 C respectively.
a) How strong is the electric field between the plates?
4kQ 4·9 x 109 Nm2/C2 · 36 x 10-6 C
E=
=
= 4.07 x 106 N/C
A
(1 m)2
b) If a 2 cm by 10 cm by 10 cm metal brick is slid between the plates with the broad sides parallel to the
two plates, how much charge will it build up on the sides of the plate? (Hint: what is the net field inside
a block of metal?)
The net field inside the block of metal will have to be zero, because it is a conductor. This happens
because charges will gather on the broad sides of the block until they create a field that exactly
cancels out the outside field. In this case enough charge will have to accumulate to create a 4.07 x
106 N/C field inside the block to cancel the one from the parallel plates. The broad sides of the
block will act like parallel plates, so we can use the parallel plate equation to calculate the amount of
charge in either side of the block:
4kQ
E=
A
EA 4.07 x 106 N/C · (0.1 m)2
Q=
=
= 3.60 x 10-7 C = 0.36 C
4k
4· 9 x 109 Nm2/C2
12. A 10 m radius, copper sphere that carries a charge of +295 C is suspended by insulating wire at
the center of a hollow, aluminum shell (inner radius 22 m, outer radius 27 m).
a) Draw a sketch of the situation, showing the charges and field lines.
b) What is the charge on the inner surface of the aluminum shell? What is the charge on the outer
surface of the aluminum shell?
The inner surface is charged to – 295 C and the outer surface is charged to +295 C.
c) How strong is the electric field at each of the following points? (All distances measured from center)
i) 5 m from center; E = 0 N/C
You are inside a conductor and the field is always 0 inside a conductor in a static
situation. Another way of thinking about is that all of the charges arranged around the
outside of the sphere create fields that all cancel each other out perfectly wherever you
are inside the sphere.
kq (9 x 109 Nm2/C2)(295 x 10-6 C)
ii) 15 m from center; E = r2 =
= 11,800 N/C
(15 m)2
Outside of the sphere, the sphere creates a field that looks exactly like the field coming
from a point charge located at the center of the sphere. To calculate the field outside of
the sphere, we pretend that it is a point charge located at the center. [The charges
arranged around the outer shell have no effect because their fields cancel out inside the
shell – you are on the inside of the conductor]
iii) 25 m from center; E = 0 N/C
You are inside a conductor and the field is always 0 inside a conductor in a static
situation. In this case we can say that the field between the inner and outer charges on
the shell exactly cancels out the field from the charged ball in the center, leaving no net
charge.
kq (9 x 109 Nm2/C2)(295 x 10-6 C)
iv) 35 m from center; E = r2 =
= 2,167.3 N/C
(35 m)2
Outside of the sphere, the sphere creates a field that looks exactly like the field coming
from a point charge located at the center of the sphere.