32: Tactics Box 32.1 Right-Hand Rule for Fields

Figure 1

Figure 2

INTRO:
TACTICS BOX 32.1 Right-hand rule for fields

1. Point your right thumb in the direction of the current.(Figure 1)
2. Curl your fingers around the wire to indicate a circle.
3. Your fingers point in the direction of the magnetic field lines around the wire.

PART A: The following sketches show a wire carrying a current I in the direction indicated. Which sketch correctly shows the magnetic field lines around the wire?
a)b)c)

SOLUTION:
First off, we know that the magnetic field can't be going two different directions, such as it is in b), which rules out that answer. Next we must use the right hand rule to determine the direction it goes. Pointing our right thumb downward, in the direction thatI points, our fingers wrap counterclockwise around the bar. Since a) demonstrates this, we know that the correct answer is a). 

PART B: The magnetic field lines around a current-carrying wire are shown in the figure. In what direction is the current flowing in the wire? (Figure 2)

SOLUTION: For this one, pay special attention to which of the blue lines are in front of the diagram of the rod and which ones are behind, because this determines the direction of the magnetic field and also how to determine the direction of the current. So, that being said, place a right hand in the air and curl your fingers in the direction of the field, towards you as if you're looking at your fingernails. This results in the thumb pointing to the right. This means that this is the direction that the current is flowing in the wire. So the final answer for this part is 'from left to right' 

29: Problem 29.9

PART A:
What is the direction of the electric field at the dot in the figure?

SOLUTION: 
placing the x axis horizontally through the point, it can be figured that V(x) = 10000x
E(x) = -dV(x)/dx so E(x) = -10000 V/m
this means that, based on our imposed axis, the direction of the electric field is to the left

PART B:
What is the magnitude of the electric field at the dot?

SOLUTION:
We already figured this out in the last part, and it is 10000 V/m

29: Problem 29.76

INTRO: The electric potential in a region of space is V=100(x²−y²) V, where x and y are in meters.

PART A:
Find an expression for the electric field E⃗ at position (x,y).
Enter your answers separated by a comma.

SOLUTION: 
The formula for this is E=-∇V
∇V=<∂V/∂x, ∂V/∂y>
∂V/∂x = 200x
∂V/∂y = -200y
E(x,y)=<-200x, 200y>

29: Problem 29.44

INTRO: The electric potential in a region of space is V=350V⋅m /√(x²+y²), where x and y are in meters.

PART A:
What is the strength of the electric field at (x,y)=(3.0m,3.0m) ?
Express your answer using two significant figures.

SOLUTION: 
350⋅(x²+y²)^-1/2
The formula for this is E=|∇V|
∇V=<∂V/∂x, ∂V/∂y>
∂V/∂x = -350x(x²+y²)^-3/2
∂V/∂y = -350y(x²+y²)^-3/2
E(x,y)=<-350x(x²+y²)^-3/2,-350y(x²+y²)^-3/2>
E(3,3) = <-175/9√2,-175/9√2>
|E| = 175/9 V/m = 19 V/m

PART B: What is the direction of the electric field at (x,y)=(3.0m,3.0m) ? 
Give the direction as an angle (in degrees) counterclockwise from the positive x-axis.

SOLUTION:
E(3,3) = <175/9√2,175/9√2>
since both sides are the same, it forms a perfect right triangle resulting in θ=45°

29: Problem 29.43

INTRO: The electric potential in a region of space is V=(180 x²− 280 y²) V, where x and y are in meters.

PART A:
What is the strength of the electric field at (x,y)=(1.0m,3.0m) ?
Express your answer using two significant figures.

SOLUTION: 
The formula for this is E=∇V
∇V=<∂V/∂x, ∂V/∂y>
∂V/∂x = 360x
∂V/∂y = -560y
E(x,y)=<360x,-560y>
E(1,3) = <360,-1680>
to get the magnitude of the vector, √(360)²+(1680)²|
E = 1718 V/m = 1700 V/m

PART B: What is the direction of the electric field at (x,y)=(1.0m,3.0m) ? 
Give the direction as an angle (in degrees) counterclockwise from the positive x-axis.

SOLUTION:
so, with the vector that we found before, E(1,3) = <360,-1680>, we can initially determine the angle that it is pointing from below the positive x axis
the 'adjacent' side of the triangle this vector creates is 360
the 'opposite' side is 1680
tan(1680/360) would solve the angle in that right triangle
θ=22°
now the question asks for the direction as an angle counterclockwise from the positive x axis, so the exact opposite way that we just determined. Since the electric field points the opposite way [E=-dV/dx], our angle isn't in the 4th quadrant it's actually in the 2nd. 
E = <-360, 1680>
θ=102°

29: Problem 29.41

PART A: 
Determine the magnitude and direction of the electric field at point 1 in the figure

SOLUTION: 
I like to start out by naming the rings
A=0V
B=25V
C=50V
D=75V
point 1 is on ring C and 1 cm away from rings B & D
A formula to determine the potential along this line would be 2500d, where d is the distance from ring A in meters
since E(x) is the negative derivative of V(x), E(x) = -2500 V/m, or in terms of vectors, E = 2500 V/m down

PART B:
Determine the magnitude and direction of the electric field at point 2 in the figure.

SOLUTION:
similar solution, except along this line the rings are much closer together, like 0.5 cm apart. 
a function for this line going towards the A ring is 5000d. Take the derivative again and it's 5000 V/m. Remember that it is also negative so in terms of vectors, it would be pointing up. E = 5000 V/m up

29: Problem 29.4

INTRO: The figure is a graph of E_x. The potential at the origin is -250 V

PART A: 
What is the potential at x = 3.0 m?
Express your answer to two significant figures and include the appropriate units.

SOLUTION: 
To determine the potential at the given point, we need to determine the potential difference and add the initial potential. V_f=ΔV + V_i
To get the potential difference we need to integrate the function in the graph 
ΔV =- ∫_xi^x_f E(x)dx 
since the function E(x) changes at 2, we need to integrate two different functions and put them together
The function when 0≤x≤2 is simply y=200
the function when 2≤x≤3 is y=-200x+600
the integral for 0≤x≤2 is - ∫₀² 200 dx
the integral for 2≤x≤3 is - ∫₂³ -200x+600 dx
Now to integrate the two of those,
1) - ∫₀² 200 dx = -200x |₀² = -400-0 = -400 V
2) - ∫₂³ -200x+600 dx = 100x²-600x |₂³
= 900-1800-(400-1200) = -900+800 = -100 V
ΔV = -400 V + -100 V = -500 V
V_f=ΔV + V_i = -500 V + -250 V = -750 V

29: Problem 29.3

INTRO: The figure is a graph of E_x

PART A: 
What is the potential difference between x_i=2.0m and x_f=3.0m?
Express your answer to two significant figures and include the appropriate units.

SOLUTION: 
To get the potential difference we need to subtract the initial potential from the final potential
ΔV=V_f -V_i
We start out by determining those two values based on the values of E_x that can be seen in the graph
E_xi = E_{2 m} = 100 V/m
E_xf = E_{3 m} = 200 V/m
ΔV =- ∫_xi^x_f E(x)dx 
for a function of E(x), y=100x-100
- ∫₂³ 100x-100 dx = -(50x²-100x) |₂³
= [-50(3)²+100(3)]-[-50(2)²+100(2)] =-450+300+200-200 = -150 V

29: Problem 29.29

INTRO: A 1.50-cm-diameter parallel-plate capacitor with a spacing of 0.700mm is charged to 500V .

PART A: 
What is the total energy stored in the electric field?
Express your answer with the appropriate units.

SOLUTION: 
First off I'm going to convert all of the units to SI units
diameter = 1.50 cm = 0.015 m
spacing = 0.700 mm = 7×10^-4 m
The energy stored in a parallel plate capacitor is determined using the following formulas
U = Q²/(2C) = (CV²)/2 = (QV)/2
We can get Q by using the formula V=(Qd)/(ε₀A) ⇒ Q = (Vε₀A)/d
A = πr² = π(diameter/2)² = 1.77×10^-4 m²
plugging all of our values into the formula for Q we get
Q = 500V ⋅ 8.85×10^-12 C/V ⋅ 1.77×10^-4 m² ⋅1/7×10^-4 m
which gives us Q = 1.119×10^-9 C
Plugging this value of Q into the formula for energy we get
U = 1.119×10^-9 C ⋅ 500 V ⋅ 1/2 = 2.798×10^-7 J 

PART B: 
What is the energy density?
Express your answer with the appropriate units.

SOLUTION: 
The formula for energy density is u= U/volume
the volume of the capacitor is A⋅spacing = 1.77×10^-4 m²⋅7×10^-4 m
giving us Vol=1.239×10^-7 m³
plugging that back into the formula for energy density,
u=2.798×10^-7 J / 1.239×10^-7 m³
= 2.258 J/m³

29: Problem 29.15

INTRO: The electric potential along the x-axis is V=100e^−2x/m V, where x is in meters.

PART A: What is E_x at x=1.0m?
Express your answer as an integer and include the appropriate units.

SOLUTION: 
E = -dV(x)/dx, the derivative of V is - 100e^−2x/m⋅(-2) V/m = 200e^−2x/m V/m
E(1) = 200e⁻² V/m = 27.067 V/m ≈ 27 V/m

PART B: What is E(x) at x=1.6m?
Express your answer to two significant figures and include the appropriate units.

SOLUTION: 
E = -dV(x)/dx, the derivative of V is - 100e^−2x/m⋅(-2) V/m = 200e^−2x/m V/m
E(1.6) = 200e^−2⋅1.6 V/m = 200e^−3.2 V/m = 8.2 V/m

29: Problem 29.14

PART A:
The electric potential along the x-axis is V =200x²V, where x is in meters. What is E_x at x =0m?
Express your answer as an integer and include the appropriate units.


SOLUTION:
E = -dV(x)/dx, the derivative of V is -400x V/m
E(0) = -400(0) V/m = 0 V/m

PART B:

What is E_x at x =3m?

Express your answer to three significant figures and include the appropriate units.

SOLUTION:
E = -dV(x)/dx, the derivative of V is -400x V/m
E(3) = -400(3) V/m = -1200 V/m = -1.20×10³ V/m

29: Potential Difference and Potential near a Charged Sheet

INTRO: Let A=(x1,y1) and B=(x2,y2) be two points near and on the same side of a charged sheet with surface charge density +σ. The electric field E⃗ due to such a charged sheet has magnitude E=σ/(2ϵ0) everywhere, and the field points away from the sheet, as shown in the diagram

PART A: 

What is the potential difference 

VAB=VA−VB between points A and B?

Express your answer in terms of some or all of E, x1, y1, x2, and y2.

  SOLUTION:

Begin by considering the line integral along the path from B to A. 

The formula for potential difference, V between the two points doesn't depend on the path, lets call it l, but rather the end points, A & B. 

VAB=−∫ABE⃗ ⋅dl⃗ ,

The line integral from B=(x2,y2) to A=(x1,y1) of C⃗ =Cxı^+Cyȷ^, a constant vector field (i.e., independent of x and y), is given by

∫ABC⃗ ⋅dℓ⃗ =∫x1x2Cxdx+∫y1y2Cydy
=Cx(x1−x2)+Cy(y1−y2).

Since the charge is consistent throughout the x direction, the x component can be disregarded

This leaves us with the solution

VAB= −E(y1−y2)

−E(y1−y2

NOTE:
the expression E(y2−y1) will not yield the correct potential if you apply it to two points on opposite sides of the sheet. For example, the expression does not indicate that two points on opposite sides of the sheet and the same distance from it are at the same potential (VAB=VA−VB=0), which is clear from the symmetry of the situation. If you take care in carrying out the integration to observe the change in the direction of the electric field as you pass from one side of the sheet to the other, you will find that the potential difference between A and B is actually given by

VAB=E(|y2|−|y1|).

PART B: 

If the potential at y=±∞ is taken to be zero, what is the value of the potential at a point VA at some positive distance y1 from the surface of the sheet?

  SOLUTION:

Substitute appropriate values for VB and y2 in the equation VA−VB=E(|y2|−|y1|).

VB = 0

y2 = ±∞

VA=E(∞) = ∞

PART C: 

Now take the potential to be zero at y=0 instead of at infinity. What is the value of VA at point A some positive distance y1 from the sheet?

  SOLUTION:

Substitute appropriate values for VB and y2 in the equation VA−VB=E(|y2|−|y1|).

VB = ±∞

y2 = 0

VA= E(0 -|y1|) 

= −E⋅y1