28: Problem 28.19

INTRO:
A 3.4-cm-diameter parallel-plate capacitor has a 1.6 mm spacing. The electric field strength inside the capacitor is 8.0×10⁴ V/m .
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PART A:
What is the potential difference across the capacitor?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

My favorite first step,​

Givens/ conversions:
diameter ≡ d = 3.4 cm = 0.034 m
spacing ≡ s = 1.6 mm = 0.0016 m
Electric field inside ≡ E = 8.0×10⁴ V/m

using equation 28.5, for the electric potential inside a parallel-plate capacitor: V = Es

so, V = (8.0×10⁴ V/m)(0.0016 m) = 128 V

the frustrating kicker here, it got me anyway, is that the instructions say to express your solution to two significant figures. 
128 has 3 sigfigs. annoying. You have to round up to 130...

⇒ ΔV_C = 130 V

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PART B:
How much charge is on each plate?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
An electric field is considered to be a surface charge density (with units of C/m²) divided by the vacuum permittivity constant (with units C/Vm). This means that the surface charge density, η, can be calculated by multiplying the electric field by the vacuum permittivity constant. 

η=E⋅ε₀​

the surface charge density is basically the charge distributed over an area, or η = Q/A
therefore, to solve for the charge, 

Q = ηA = E⋅ε₀⋅A
​

The area is just the area of the circular plate, or A = πr² = π/4 ⋅ d²

∴ Q = E⋅ε₀⋅π/4⋅d²

= (8.0×10⁴ V/m)(π/4)(8.854×10^-12 C/Vm)(0.034 m)²
⇒ Q = 6.431×10^-10 C​

but don't forget sigfigs... it only wants two...

⇒ Q = 6.4×10^-10 C​

28: PSS 25.2 The Electric Potential of a Continuous Distribution of Charge

INTRO:
To practice Problem-Solving Strategy 25.2 for continuous charge distribution problems.
A straight rod of length L has a positive charge Q distributed along its length. Find the electric potential due to the rod at a point located a distance d from one end of the rod along the line extending from the rod. (Figure 1)

Figure 1

PROBLEM-SOLVING STRATEGY 25.2 The electric potential of a continuous distribution of charge
MODEL: Model the charges as a simple shape, such as a line or a disk. Assume the charge is uniformly distributed.
VISUALIZE: For the pictorial representation:
Draw a picture and establish a coordinate system.
Identify the point P at which you want to calculate the electric potential.
Divide the total charge Q into small pieces of charge ΔQ using shapes for which you already know how to determine V. This division is often, but not always, into point charges.
Identify distances that need to be calculated.
SOLVE: The mathematical representation is V=ΣVi.
Use superposition to form an algebraic expression for the potential at P.
Let the (x,y,z) coordinates remain as variables.
Replace the small charge ΔQ with an equivalent expression involving a charge density and a coordinate, such as dx, that describes the shape of charge ΔQ. This is the critical step in making the transition from a sum to an integral because you need a coordinate to serve as an integration variable.
Express all distances in terms of the coordinates.
Let the sum become an integral. The integration will be over the coordinate variable that is related to ΔQ. The integration limits for this variable will depend on the coordinate system you have chosen. Carry out the integration, and simplify the result.
ASSESS: Check that your result is consistent with any limits for which you know what the potential should be.

Model
No information is provided on the rod's cross section. For simplicity, assume that its diameter is much smaller than the rod's length. You can, then, model the rod as a line of charge.

Visualize
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PART A:
To most efficiently solve this problem, you should divide the rod into pieces of charge that consist of...

- thin lines of charge of length L  and a very small cross section
- thin 'slices' of the rod cut parallel to the axis of the rod
- thin 'slices' of the rod cut perpendicular to the axis of the rod

SOLUTION:
First, model the rod as a line of charge. Then, divide the line into many small segments, each of length Δx. Each segment can, then, be modeled as a point charge. The potential due to such a point charge can be determined in a relatively straightforward manner.

∴ option #3, thin 'slices' of the rod cut perpendicular to the axis of the rod
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PART B:
Choose a coordinate system where +x is to the right and +y is upward. Place the origin of your coordinate system at the left end of the rod, and choose point P to be located beyond the right end of the rod, as shown in the picture to the left.
What is the distance ri between point P and a piece of charge located at position xi?
Express your answer in terms of some or all of the quantities xi, L, Q, and d.

SOLUTION:
A pictorial representation for this problem might look like:

r_i = (L+d)-x_i
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PART C:
Find the electric potential V_P at point P.
Express your answer in terms of d, L, Q, and ϵ₀.

SOLUTION:
The mathematical expression for the potential is V=ΣVi. 
Because of the nature of the problem, which deals with a continuous distribution of charge rather than a point-like distribution, the sum must be treated as an integral. 

To set up the summation and find what to integrate, follow the steps listed in the strategy ^^

After dividing the total charge into small segments, and modeling each segment as a point charge, as in Part A, construct a mathematical expression for Vi, the electric potential due to segment i. 
Treat xi, the x coordinate of segment i, as a variable (just call it x), and express the charge on that segment in terms of said x. 

Finally, let ΣVi become an integral and use calculus to evaluate the integral...

FIND... the electric potential Vi at point P due to the charge segment located at the variable position 
x

The formula for the electric potential of a point charge involves the distance from the charge to the point where the potential is to be computed.
Recall that, in Part B, you obtained an expression relating the distance between point P and charge segment i to the x coordinate of the charge segment.......

The formula for electric potential is... 
U_elec = k q₁q₂/r
and we all know what k is so that will be temporarily ignored...

The immediate distance between the two points is r_i, as determined in part B... 

U = k q₁q₂/r ...  then just plug in for r as r_{i ...}
V_i = k ⋅ΔQ/(L+d-x_i)

Next you need to understand the expression for the charge of a small segment
So calculus is kind of annoying at first, but these topics are actually really important to understand. especially later on (trust me)..

Because you will integrate with respect to coordinate x, it is important to express the charge on a segment, ΔQ, in terms of its length Δx.

So you gotta find a mathematical expression for ΔQ, while assuming the charge is uniformly distributed along the rod (since that's given). 

Recall that the total charge on the rod is Q and the rod's length is L...
that means that the change in charge over the change in x is simply the charge divided by the length, or...
ΔQ/Δx = Q/L
by rearranging this to solve for ΔQ...

ΔQ = Δx⋅Q/L

plug that into your initial V_i equation for ΔQ, 
V_i = k ⋅Δx⋅Q/L ⋅/(L+d-x_i)
you now have an integrable function..
V_i = ∫k⋅Q/L⋅(L+d-x_i) Δx = ∫k⋅Q/L⋅(L+d-x) dx
integrate from 0 to L, because the d is already attributed... 

k is constant, Q is constant, L is constant, so you can just pull that out...
→k⋅Q/L ⋅∫1/(L+d-x) dx
use substitution...

let u = L+d-x, ∴du = -dx

→k⋅Q/L⋅∫1/(u) -du
→k⋅Q/L⋅-ln(u) (from 0 to L)usub
→k⋅Q/L⋅-ln(L+d-x) (from 0 to L)
→k⋅Q/L⋅[-ln(L+d-L) - -ln(L+d-0)]
→k⋅Q/L⋅[-ln(d) + ln(L+d)]
using law of logs, subtracting them is the same as dividing them...
→k⋅Q/L⋅ln[(L+d)/d] = k⋅Q/L⋅ln[L/d +d/d]

⇒V_p = k⋅Q/L⋅ln[1+ L/d] = Q/4Lπε₀⋅ln[1+ L/d]

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PART D:
Imagine that distance d is much greater than the length of the rod. Intuitively, the potential should be approximately the same as the potential at a distance d from which of the following charge distributions?

- an infinitely long wire with total charge Q
- an infinitely long wire with total charge Qd/L
- a point charge of magnitude Q
- an electric dipole with moment QL

SOLUTION:
From far away, a short rod looks very much like a simple point charge. Not surprisingly, the mathematical expression you obtained for the potential does reduce to that of a point charge if L/d≪1:

In other words, since L is constant, the larger d is, the smaller L/d gets. When L/d is really small, the logarithm is basically just the logarithm of 1 plus that tiny little amount. The ln(1) = 0, and oddly enough, at really close proximity to 1, those values follow the rule :

if x << 1 & ln(1+x), then ln(1+x) ≈ x​

try this with x = 0.001, for example... ln(0.001) = 0.0009995 ≈ 0.001

therefore, Q/4Lπε₀⋅ln[1+ L/d] where L/d << 1 ≈ Q/4Lπε₀⋅L/d 
⇒ or, ≈ Q/4dπε₀

Therefore, the third option is correct. or, 

a point charge of magnitude Q​

28: PSS 25.1 Conservation of Energy in Charge Interactions

INTRO:
A proton and an alpha particle are momentarily at rest at a distance r from each other. They then begin to move apart. Find the speed of the proton by the time the distance between the proton and the alpha particle doubles.
Both particles are positively charged. The charge and the mass of the proton are, respectively, e and m. The charge and the mass of the alpha particle are, respectively, 2e and 4m.

PROBLEM-SOLVING STRATEGY 25.1 Conservation of energy in charge interactions
MODEL: Check whether there are any dissipative forces that would prevent the mechanical energy from being conserved.
VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find.
SOLVE: The mathematical representation is based on the law of conservation of mechanical energy:

K_f+qV_f=K_i+qV_i.

Is the electric potential given in the problem statement? If not, you'll need to use a known potential, such as that of a point charge, or calculate the potential using the procedure given in Problem-Solving Strategy 25.2.
K_f and K_i are the sums of kinetic energies of all moving particles.
Some problems may need additional conservation laws, such as conservation of charge or conservation of momentum.
ASSESS: Check that your result has the correct units and significant figures, is reasonable, and answers the question.

Model
The particles described in this problem interact under the effect of the electric force, which is a conservative force, so the system's mechanical energy is conserved.

Visualize
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PART A:
Which of the following quantities are unknown?
A. initial separation of the particles
B. final separation of the particles
C. initial speed of the proton
D. initial speed of the alpha particle
E. final speed of the proton
F. final speed of the alpha particle
G. mass of the proton
H. mass of the alpha particle
I. charge of the proton
J. charge of the alpha particle

Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if A, C, and D are unknowns, enter ACD.

SOLUTION:
The final separation between the particles is, essentially, known: It is twice the initial separation, or 2r. In this problem, then, there are really only two unknowns: the final speed of the proton, (v_f)_p, which is what you are trying to find, and the final speed of the alpha particle, (v_f)_α.

Recall that a problem with two unknowns requires two equations to be solved. Here, the law of conservation of mechanical energy provides one of the equations. To find the second equation, think what other physical quantity besides energy is conserved, and translate that into a mathematical expression. But before you do that, it's helpful if you complete a before-and-after pictorial representation of the problem. Your drawing might look like this:

answer: BEF
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PART B:
Find the speed of the proton (v_f)_p by the time the distance between the particles doubles.
Express your answer in terms of some or all of the quantities e, m, r, and ϵ₀.

SOLUTION:
Using the law of conservation of mechanical energy, we know that the sum of the initial energies is equal to the sum of the final energies. 

At first, when the two particles are right next to each other, there is no kinetic energy yet, but quite a bit of potential energy.  
The potential energy can be written using U = qV -> U_elec = 1/[4π⋅ϵ₀]⋅q₁q₂/dist

Therefore, since initial dist = r, q₁ = e & q₂ = 2e
The initial potential energy = 
U_i = 1/[4π⋅ϵ₀]⋅2e⋅e/r = 1/[4π⋅ϵ₀]⋅2e²/r

and sine the final dist = 2r,

the final potential energy = 

U_f = 1/[4π⋅ϵ₀]⋅2e⋅e/r = 1/[4π⋅ϵ₀]⋅e²/r

What about the kinetic energy? 

if the particles start at rest, then the initial kinetic energy = 

K_i = 0

the total final kinetic energy will be equal to the final kinetic energy of the proton plus the final kinetic energy of the alpha. But how can we relate these two velocities to each other? 

The only quantity that remains constant for the proton and alpha as they move apart is their magnitude of momentum. 

momentum : P = m*v

conservation of momentum: ∑P₁ = ∑P₂
∴ v_α1⋅m_α + v_p1⋅m_p = v_α2⋅m_α + v_p2⋅m_p
since both initial velocities are 0, 
0 = v_α2⋅m_α + v_p2⋅m_p
→v_α⋅m_α = v_p⋅m_p

→v_α/v_p = m_p/m_α = m/4m = 0.25

∴ v_α = (0.25)v_p​

KE = 1/2 mv²
∴ KE = 1/2m_αv_α² + 1/2m_pv_p² 
sub in for v_α = (0.25)v_p...
KE = 1/2(4m)((0.25)v_p)² + 1/2(m)v_p² 

∴KE = (5/8)mv_p²​

plugging that into the final equation... 
0 + 1/[4π⋅ϵ₀]⋅2e²/r = (5/8)mv_p² +1/[4π⋅ϵ₀]⋅e²/r
then solve for v!
→ 1/[4π⋅ϵ₀]⋅e²/r = (5/8)mv_p² 
→ 2/[5π⋅ϵ₀]⋅e²/r = mv_p² 
⇒∴v_p = SQRT{2⋅e²/[5r⋅m⋅π⋅ϵ₀]}
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PART B:
The best way to check whether your result from Part B is correct is to check that it has the correct units. Which of the following expressions, where C stands for coulombs, N for newtons, kg for kilograms, and m for meters, represents the correct SI units for the expression found in part B?


SOLUTION:
just plug in for the units... 
e ≡ C, r ≡ m, m ≡ kg, ϵ₀ ≡ F/m = C²/Nm²
and all numbers have no units... 
→ v = SQRT{C²/[m⋅kg⋅C²/Nm²]]}
The C's cancel out, one of the m's cancel out, and the N jumps up to the numerator...

⇒ units = √N⋅m/kg

28: Energy Stored in a Charge Configuration

INTRO:
Four point charges, A, B, C, and D, are placed at the corners of a square with side length L. Charges A, B, and C have charge + q, and D has charge − q. (see Figure 1) 

Throughout this problem, use k in place of 1/(4πϵ₀).

Figure 1	Figure 2

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PART A:
If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinitely far apart, you will find that the contribution for each charge is proportional to kq²/L. In the space provided, enter the numeric value that multiplies the above factor, in W.

SOLUTION:
We know that the work to assemble a charge configuration of two charges a distance r from each other is simply W = kq²/r
If we want to assemble three charges A, B, and C... you have to consider the distances between all of them
W_ABC = kq²/(r_AB + r_AC + r_BC)
finally, to assemble four charges A, B, C, & D....
W_ABCD = kq²/(r_AB + r_AC + r_AD + r_BC + r_{BD + rCD})

Note, however, that this is for all positive charges. If there is a negative charge in there, you must subtract that distance rather than add it to the divisor. 

Considering a square charge configuration with sides L, such as in figure 1... A, B, & C are positive & D is negative
r_AB = L
r_AC = L√2
r_AD = L (-)
r_BC = L
r_BD = L√2 (-)
r_CD = L (-)

⇒ W = kq²/(L + L√2 + (-L) + L + (-L√2) + (-L)
⇒ ∴ W = 0 × (kq²/L)
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PART B:
Which of the following figures depicts a charge configuration that requires less work to assemble than the configuration in the problem introduction? Assume that all charges have the same magnitude q. (Figure 2)

SOLUTION:
working through each option... 

(a)
This one appears to be complicated at first, but really isn't if you consider trig. 
The positive charges are equidistant from each other at a distance of L. This means that the negative charge is as far away from them as half of the distance between one of the positive charges and the corresponding line opposite it. 
r_AB = L
r_AC = L
r_AD = ½L⋅sin(60) (-)
r_BC = L
r_BD = ½L⋅sin(60) (-)
r_CD = ½L⋅sin(60) (-)
Wa = kq²/(3L - (3/2)L⋅(0.866))
⇒ ∴ Wa = (1/1.7) × (kq²/L) = (0.5879)× (kq²/L)

(b)
r_AB = L
r_AC = 2L
r_AD = 3L (-)
r_BC = L
r_BD = 2L (-)
r_CD = L (-)

Wb = kq²/(4L - 6L)
⇒ ∴ Wb = (-1/2) × (kq²/L) = (-0.5)× (kq²/L)

(c)
This one is identical to the problem introduction problem, but multiplied by a factor of 2. The factor doesn't matter, so Wc = 0 × (kq²/L)

In this case, the greater work is actually the less work. By this I mean, the positive work represents the amount of work the system actually exhibits, that we don't have to do. If there is negative work, we have to make up that work in order to place the charges as desired. 
Therefore, charge configuration (a) requires the least amount of work.

figure a