Notes: Uniform Fields and Capacitance
Level 4
So far in this unit, we’ve talked about fields and charges caused by a few particles. These create fields that change
depending upon your position near them. In this section, we are going to explore uniform electric fields- or, fields that
are pretty much the same in any given area.
Creating Uniform Fields
It’s actually a pretty elegant idea. All you need to do this is essentially two metal plates, a battery,
and wires to connect to the battery. Sometimes, we stick an insulator between the plates as well.
We bring them very close together (so that the distance between them is minor compared to their
size, not like in the picture) and ta da! a uniform field is born between them.
We call this particular set up- charged plate, insulator, charged plate- a capacitor. How does this
set up create a uniform field? Well a battery is essentially a device that will steal electrons from
the positive side and pump them over to the negative side. When we hook the plates up, this is
exactly what happens- the battery yanks negative charge from the plate hooked up to the positive
terminal and pumps them onto the plate hooked up to the negative charge.
We now have a sheet of negative charges and a sheet of positive charges that sit facing one
another. Think about it- what would happen if we put a test charge (remember, test charges are
positive) between these two- where would it want to go? Which direction would the electric field
point?
It would want to travel away from the positive plate and toward the negative plate. But how can this be uniform? Won’t
the particle be more attracted to the negative plate when it is closer? Yes, it will be further away from the positive plate,
which was also helping to push it in that direction. If we move the test particle close to the positive plate, it won’t have
the negative plate pulling on it as much, but it will be feeling a big repulsive push from the positive plate. These forces
constantly balance, create a uniform field between the plates.
One important small point to keep in mind- see how the electric field bubbles out a little on the edges in the picture on
the right? We call that edge effect, and you will normally be told to ignore it. Just sayin’.
Describing Capacitors
There are several ways we can mathematically describe capacitors. One is to think about the way the capacitor is
actually made. We describe a capacitors capacitance by using physical characteristics of the capacitor- namely, the area
of it’s plates and their separation.
from Princeton Review
The section above only applies to parallel plate capacitors. Will you have to do calcuations with different types? No. The
equation that applies to all capacitors is this:
𝐶=
𝑄
∆𝑉
Where Q is the charge that is able to build up on a plate (that is stored by the capacitor) and ∆𝑉 is the difference in
potential between the two plates. This is sometimes abbreviated as V because it is the same as the voltage on the
battery.
Practice Problem: Calculating with Capacitance
Solution
Practice Problem: Charge on each Plate
Solution
Occasionally, we also describe capacitors using their charge density. We use the symbol 𝜎 for this. It is, very simply, the
amount of charge per area on a plate. If we know the charge on one of the plates, we can calculate the electric field
between them using the equation:
𝐸=
𝜎
𝜀𝑜
Potential and Potential Energy in Capacitors
On the last level, we discussed how much energy it took to move a particle in an electric field from one position to
another. How do we do that in a uniform magnetic field? In some ways, this is actually easier. We use the equation:
𝑊 = 𝐹𝑑
Where F is the force, and d is the distance you move the charge. In a uniform electric field, the E is constant, so we can
now use the equation:
𝐹
𝐸 = 𝑞 or 𝐹 = 𝐸𝑞
Plugging this in, we get,
𝑊 = 𝐸𝑞𝑑
Hold that thought for a second. We are going to come back to it. In the previous section, we also learned that when we
drag an object from point A to point B:
𝑊 = ∆𝑈 = 𝑈𝐵 − 𝑈𝐴
We should also recall that
𝑉=
𝑈
𝑞
Therefore:
𝑊 = ∆𝑈 = ∆𝑉𝑞
When we talk about the parallel plates, what causes the potential difference between them? The battery. So the
potential difference between the plates is the same as the voltage on the battery. Because of this, we often abbreviate
∆𝑉 as V, because we are talking about the voltage on the battery.
𝑊
=𝑉
𝑞
Now, let’s return to the equation we found before:
𝑊 = 𝐸𝑞𝑑
If we combine this with the other equation, we get:
𝐸𝑞𝑑
=𝑉
𝑞
Which is simplified to:
𝑉 = 𝐸𝑑
Where V is the voltage difference between the plates (which is the same as the voltage on the battery) and d is the
distance we are moving the particle in the uniform electric field. Many, many problems will ask us about moving a
charge from one plate to another, in which case d is the distance between the plates.
I really like this equation, even though it isn’t often covered in textbooks, because it lets use easily calculate the electric
field between two plates, as long as we know how far apart the plates are and the voltage on the battery. It also allows
you to calculate how much work is being done on a particle as it moves from one position to another in a uniform field
(as long as you keep 𝑉 =
𝑊
𝑞
in mind as well.
Practice Problem: Work in a Uniform Field
A positive charge of 3.0 × 10–8 coulomb is placed in an upward directed uniform electric field of 4.0 × 10 4 N/C. When the charge is
moved 0.5 meter upward, the work done by the electric force on the charge is
(A) 6 × 10–4 J (B) 12 × 10–4 J (C) 2 × 104 J (D) 8 × 104 J (E) 12 × 104 J
Solution
Answer: A
W = Fd = qEd
Practice Problem
A 5 × 10–6 coulomb electric charge is placed midway between two parallel metal plates connected to a 9–volt battery. If the electric
charge experiences a force of 1.5 × 10–4 newtons, what is the separation of the metal plates?
(A) 6.75 × 10 –9 m (B) 2.7 × 10–4 m (C) 3.7 × 10–3 m (D) 0.30 m (E) 3.3 m
Solution
Answer: D
F = Eq and E = V/d giving d = qV/F
Practice Problem: Energy in motion
A potential difference V is maintained between two large, parallel conducting plates. An electron starts from
rest on the surface of one plate and accelerates toward the other. Its speed as it reaches the second plate is
proportional to
Solution
Answer: C
W = K = qΔV and K = ½ mv2
A word about how potential varies in a capacitor. The electric field strength is constant in a capacitor, which means that
for any given charge, the force acting on it will be pretty much constant between the plates. This is very similar to the
gravitational field close to the earth- the field pulling us down is pretty steady, though larger masses will experience
more force acting on them. We can store energy in our gravitational field by lifting an object up off the ground- how
much depends upon the position and the mass. We can similarly release that energy by letting go of the object and
watching it fall. This is pretty much the same in capacitors- except the amount that is stored depends upon the charge
and the position.
You do need to keep you wits about you when solving these, however, because charges that have the same magnitude
can feel different potentials depending on whether they are positive or negative.
For positive, charges, this is easy enough. Think about a mass in a constant
gravitational field (like the one you are sitting in right now). The mass is attracted to
the ground, so pulling it away from the ground is essentially storing energy in the mass.
Which plate is the positive charge attracted to? If you said the negative one, good- you
aren’t sleeping. So if we pull the positive charge away from the negative plate, it will
gain potential energy. This is a position of higher potential. The further it is away, the
more potential it stores. For a positive charge, the electric field lines point in the
direction of lower potential energy. This is actually the convention that we stick with, too. If we were to hook the two
plates in the picture up to a 6 V battery, we would say that the positive plate had 6V and the lower (negative plate) had
0 V.
Which plate is the negative charge attracted to? The positive one. Therefore, pulling the charge away from the positive
plate will cause it to gain potential energy. It seems strange, but even though the electron is at the position with (by
convention) 0V, it has a lot of potential and potential energy- exactly 6V of potential.
2002B5B (Modified). Two parallel conducting plates, each of area 0.30 m2, are separated by a distance of 2.0 × 10 –2 m
of air. One plate has charge +Q; the other has charge –Q. An electric field of 5000 N/C is directed to
the left in the space between the plates, as shown in the diagram above.
a. Indicate on the diagram which plate is positive (+) and which is negative (–).
b. Determine the potential difference between the plates.
An electron is initially located at a point midway between the plates.
d. Determine the magnitude of the electrostatic force on the electron at this location and state its
direction.
e. If the electron is released from rest at this location midway between the plates, determine its speed just
before striking one of the plates. Assume that gravitational effects are negligible.
Solution
e. Here is how I would solve this. First, I’d use ∆𝑉 = 𝐸𝑑 . We are talking about releasing a particle and having it move
through have the distance between the plates, so that give me my d (1.0 x 10-2). The electric field strength is 5000 N/C
(that’s my E). By plugging these in, I can find the difference in potential between the middle of the plates and a plate.
Using this, we get:
∆𝑉 = 𝐸𝑑
∆𝑉 = 50𝑉
That tells us about the potential at two different positions. It doesn’t tell us about how much energy a specific charge
would have stored if put directly between the plates. For that, we need to go to the more specific equations for potential
energy.
∆𝑉 =
∆𝑈
𝑞
We can use this to find the potential energy of charged particle half way between the plates. Plugging it in, you find ∆𝑈 =
8 𝑥 108 𝐽. Now, it is a matter of putting it into the most useful equation ever, the conservation of energy equation.
𝐾𝑖 + 𝑈𝑖 + 𝑊 = 𝐾𝑓 + 𝑈𝑓
This is no outside forces acting, so no work. There is not initial kinetic energy and no final potential (it will all be turned
into kinetic, just like a ball falling to the floor).
𝑈𝑖 = 𝐾𝑓
𝑈=
𝑚𝑣 2
2
Solve for v and you’ll find: 4.2 𝑥 106 m/s.
from Princeton Review
Shooting a Charge Through Parallel Plates
We covered this a ton in magnetism, but we might as well hit it yet again.
Most AP problems don’t just have us watching a particle accelerate from rest. Instead, they like to have us shoot the
particle in between the two plates.
Is the particle that followed this path
positively or negatively charged?
It might not look like it, but this is basically a horizontal projectile problem.
First, let’s talk about what happens to charged particles if they move perpendicular to an electric field.
from The People’s Education
So, if a particle moves from Point A to Point B, there is no change in its potential. This is the equivalent with holding a
book up and moving it back and forth without changing its height. The potential energy doesn’t change. Notice, however,
that we need to hold the book up. If we were to toss the book forward- two things would happen at the same time. In the x
direction (horizontally) the book wouldn’t experience a change in energy- it would keep going at pretty much the same
velocity (barring air resistance). In the vertical direction, however, the potential energy would start to convert into kinetic
energy. And we have ourselves a straight forward horizontal projectile problem.
It is a horizontal projectile problem, except the force acting on the particle isn’t the acceleration due to gravity, but it is the
acceleration due to the electric force. How do we find that? Easy enough.
𝐹𝑦 = 𝐹𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 = 𝐸𝑞
so
𝐹𝑦 = 𝐸𝑞 = 𝑚𝑎𝑦
Now you can solve it like a horizontal projectile problem.
Important to note: You need to keep an eye on which direction the force will act. A positive charge, for example, will be
accelerated towards a negative plate. A negative charge (like an electron, which often features in these type of
problems) will be accelerated toward the positive plate. Remember that we solve horizontal projectiles using the table?
Don’t memorize what’s below- understand where the values come from.
Horizontal (x)
Vertical (y)
𝐹𝑥 = 0
𝑥
𝑣=
𝑡
𝑘𝑄𝑞
= 𝑚𝑎
𝑟2
𝑣𝑜,𝑦 = 0
𝐹𝑦 = 𝐸𝑞 =
Practice Problem: Charges Between Plates
1985B3. An electron initially moves in a horizontal direction and has a kinetic energy of 2.0 × 103 electron–volts when it is
in the position shown above. It passes through a uniform electric field
between two oppositely charged horizontal plates (region I) and a field–free region (region II) before
eventually striking a screen at a distance of 0.08 meter from the edge of the plates. The plates are 0.04
meter long and are separated from each other by a distance of 0.02 meter. The potential difference
across the plates is 250 volts. Gravity is negligible.
a. Calculate the initial speed of the electron as it enters region I.
b. Calculate the magnitude of the electric field E between the plates, and indicate its direction on the diagram above.
c. Calculate the magnitude of the electric force F acting on the electron while it is in region I.
d. On the diagram below, sketch the path of the electron in regions I and II. For each region describe the
shape of the path.
Solution
Equipotential Lines
from Cliffs
from Princeton Review
Equipotential Lines in a Capacitor
Using Equipotential Lines to Figure out the Electric Field
This is surprisingly useful and it can be very easy. All you need to do is keep in mind two key rules:
1. Electric field lines point in the direction of decreasing potential. Remember, electric potential is scalar, so
positive values are greater than negative values.
2. Electric field lines are perpendicular to equipotential lines.
It makes a lot more sense when you see these in action.
Ms. Twu: Equipotential Surface Problem: Video 24