Download Electrostatic Forces and Energy

Electrostatic Forces and Energy
Objective: To calculate electrostatic force, electric field vector, electric potential energy
and electric potential for a given continuous or point charge distribution on a field point
that may or may not contain a charge.
Discussion: There are many relationships in static electricity that are of importance.
Electrically charged particles exert forces on other charged particles by creating electric
fields that alter their surroundings. These fields can do work on charged particles that
can be measured for a given charge. The amount of energy a point source distribution
can provide to a charged object can be determined specifically for that charged object
(electric potential energy) or in general normalized by dividing by the amount of charge
(electric potential). All of this information can be determined simply using Coulomb’s
Law for electrostatic forces and the definition of electric field, electric potential energy
and electric potential.
Governing Equations:
kq q
FES  12 2  Eq0
1.1
r
kq
E  2 rˆ
1.2
r
q
PEE  kq0  i
1.3
ri
PEE
V
1.4
q0
1.5
R  Rx2  Ry2  Rz2
1.6
rˆiˆ 
Ex ˆ
i
E
X circle  r cos 
1.7
rˆˆj 
Ey
E
ˆj
rˆkˆ 
Ycircle  r sin 
Ez ˆ
k
E
Setting Up: Open Microsoft Excel®
1. Set up your spreadsheet exactly as you see below.
A
B
C
D
E
F
G
H
I
1 K (Nm /c )
Major axis (m) Minor axis (m)
2
9e9
3
4
Source Points
5 Count
Charge (Coul) X coordinate (m) Y coordinate (m) Z coordinate (m) Rx (N/C) Ry (N/C) Rz (N/C) Potential (v)
2
2
J
K
1
2
3
4
5 Sum Rx
6 Sum Ry
7 Sum Rz
8
9
10
L
M
N
O
P
j
k
Field Point
Charge (Coul)
X Coordinate (m)
Y Coordinate (m)
Z Coordinate (m)
Direction
Magnitude
i
Electrostatic Force (N)
Electric Field (N/C)
Electric Potential Energy (J)
Potential (V)
2. The first cell will be used to count numbers in order to help write formulas to
make continuous charge distributions. Enter 0 in cell A6, 0.01 in cell A7 and 0.02
in cell A8. Highlight these cells and drag down to cell A634 to count up to 6.28
which will help make a complete circle using polar coordinates on the XY
coordinate plane.
3. Columns B, C, D, and E along with M2-5 will be data entry cells in which you
will enter the source points and the field point information. Leave these cells
empty for now.
4. In cells F6, G6, and H6 you will calculate the X, Y, and Z components of the
electric field. To do this type “=IF(B6=0,0,$A$2*B6/((C6-$M$3)^2+(D6$M$4)^2+(E6-$M$5)^2)^(3/2)*($M$3-C6))” into cell F6 and
“=IF(B6=0,0,$A$2*B6/((C6-$M$3)^2+(D6-$M$4)^2+(E6$M$5)^2)^(3/2)*($M$4-D6))” into cell G6. For the Z component type
“=IF(B6=0,0,$A$2*B6/((C6-$M$3)^2+(D6-$M$4)^2+(E6$M$5)^2)^(3/2)*($M$5-E6))” into cell H6. The first IF function is required so
that the unused points will not be divided by empty cells (zero). Now drag each
formula down 630 cells so that your spreadsheet can accommodate up to 630
three dimensional source points.
5. The electric potential for each point can be calculated in column I using Equation
1.3 and 1.4 specifically for each point charge. Enter
“=IF(B6=0,0,$A$2*B6/((C6-$M$3)^2+(D6-$M$4)^2+(E6-$M$5)^2))” into cell
I6. Note that this is cell I6 not the number 16. Again the IF function is used to
prevent from dividing by zero. Double click the formula box.
6. The sum of the X and Y components of the electric field vector can be determined
in cells H5 and H6. Enter “=SUM(F6:F636)” into cell K5, “=SUM(G6:G636)”
into cell K6, and “=SUM(H6:H636)” into K7.
7. The magnitude of the electric field vector can be calculated in cell J8 using
Equation 1.5. Enter “=SQRT(K5^2+K6^2+K7^2)” into cell M8. The direction of
the electric field vector depends on the direction of the electrostatic force on a
positive test charge at that location, so that will be calculated after the direction of
the force.
8. The magnitude of the electrostatic force can be calculated in cell M7 using the
following formula from Equation 1.1: “=M2*M8”. After typing the formula
place the curser between the M and the 2 in the formula and press the F4 key on
the keyboard. This is a quick way to switch it to a fixed reference.
9. Since electric potential energy and electric potential are scalar quantities, they do
not require a direction. Enter the following formula into cell M9 to calculate
electric potential energy: “=M2*SUM(I6:I636)”. Note that this is cell I6 not the
number 16
10. Determining direction: The unit vector with the appropriate direction can be
determined using the resultants calculated in cells K5-7. The X, Y and Z
components of the unit vector can be calculated in Columns N, O and P
respectively. Type “=$K$5/SQRT($K$5^2+$K$6^2+$K$7^2)” in cell N8,
“=$K$6/SQRT($K$5^2+$K$6^2+$K$7^2)” in cell O8 and
“=$K$7/SQRT($K$5^2+$K$6^2+$K$7^2)” in cell P8. The electrostatic force
can be in the direction of the field if the field point is positive and opposite the
direction of the field if the field point is negative. Write an if function in cells
N7-P7 to account for this. Enter “=IF($M$2<0,-N8,N8)” into cell N7 and drag it
to the right.
11. Calculate electric potential by summing the potentials of each point charge in cell
M10 by typing the following formula: “=SUM(I6:I636)”
12. Make a graph of all the X and Y coordinates using an XY scatterplot including up
to row 634. This will display your charge distribution placed in the XY Plane.
13. To start place a single point charge of 1.0 microcouloumbs on (1,1,1)
Development Questions:
1. Why is the IF function used in the formulas for Rx, Ry, Rz and Potential?
2. What do the following functions mean? A. =SQRT() B. =SUM() C. ATAN()
D. PI().
3. Quick test: What should be the magnitude of the direction vector (ijk) for the
force and electric field? Write a formula in cells Q7 and Q8 to test and record the
formula and its result below.
2 Dimensional Questions: Answer questions on a separate sheet of paper.
1. Place a charge of 1 microcoulomb (Hint: type “1e-6” to simplify) on 0,0 as a
source point and then place a charge on the field point of 1 microcoulomb on
(1,0,0).
A. Predict the field and the force for a field point at (2,0) then test.
B. Do the same for (3,0).
C. What happens to electrostatic force and electric field if the charge of the
source point is doubled?
D. What if the charge of the field point is doubled?
E. What if the field point has an opposite charge?
F. What if the source point has an opposite charge?
2. Two point charges of 10.0 microcoulombs and -5.0 microcoulombs are located
at (1.5,0,0) m and (-1,0,0) m respectively. What location (on the X axis) should
a 1 microcoulomb charge be placed in order to have zero net force acting on it?
Use goal seek and setting Electric Field to 0 by changing the X coordinate of the
field point (the Y coordinate should be zero). Hint: use -2 as a starting value
for goalseek.
3. Two spheres that can be treated as point charges are separated by a distance of
0.0500 m and have accumulated opposite charges. The electric field halfway
between them is 30,000 N/C. Calculate the charge on each sphere. Hint: Make
the charge on the second sphere equal to that of the first using a formula and
then use goal seek.
4. Four point charges with charges of 3, 4, 5, and -6 microcoulombs are located at
(0,1), (1,0), (-1,0) and (0,-1) respectively. What is the electric field magnitude,
electrostatic force, electric potential energy and voltage at the origin? The field
point has a charge of 1.00 microcoulombs.
2 Dimensional AP Questions:
1. Electric potential energy is path independent. If two positive point charges of 4.0
microcoulombs are present at (1,0) and (0,0), how much kinetic energy is gained
by a positive point charge of 2.0 microcoulombs that is placed at rest and released
at (2,0) and allowed to accelerate to (5,0)? What is the mass of the point charge if
it has a velocity of 5.0 m/sec when it reaches (5,0)?
2. Two positive point charges of 100 microcoulombs are on the X axis at a position
of (1,0) and (0,0) and there are two negative 100 microcoulomb point charges
placed at (1,5) and (0,5). What is the kinetic energy of a particle with a charge of
10 microcoulombs that is released from rest at point (0.5,0) as it passes point
(0.5,5)? Calculate the mass that the charge must have to attain a velocity of 50.0
m/sec over that interval. If a proton underwent the same displacement, what
would its velocity be? (look up the mass and charge of an electron and proton)
3. A proton is fired directly at a uranium nucleus at 8.0*107 m/sec . The nucleus is
large enough to be considered at rest throughout the problem. How close to the U
nucleus does the proton reach if the atomic number of the nucleus is 92. Assume
the effects of the electrons are negligible. Use manual guess and check since goal
seek does not like very small numbers. (your answer should be given in
nanometers)
Continuous Charge Distributions:
The superposition principle can be used to analyze continuous charge distributions. This
can be done by breaking a continuous charge up into a large number of point charges
very close together. To do this choose a large number of points (more than 100) and
space them uniformly using a formula for the charge and the X and Y coordinates. See
the answer key for specific instructions.
4. Using a 20.0 meter long line of charge along the Y axis centered on the origin
with charge per unit length 1.5*10-10 C/m, find the location on the X axis at which
the electric field is 2.5 N/C. Assume the line of charge can be replaced with 400
point charges
This is a continuous charge distribution. Therefore the distribution has to be broken up
into a large number of point charges spaced close together compared to the distance
from the distribution. In order to do this we will break up the line of charge into 400
small increments along the Y-axis centered at the origin. To do this divide total charge
(1.5*10-10*20) by 400 and place that result in cells B6 through B406.
3 Dimensional Questions:
Setting Up:
In order to place a ring of charge polar coordinates can be used. Column A counts the
angle in radians that can be used and Equations 1.7 can be used to determine the X and Y
coordinates of the point charges that will create the continuous charge distribution. Use
the following formula “=$C$2*COS(A6)” in cell C6 and “=$C$2*SIN(A6)” in cell D6 to
create the circle. Double click the formulas and a ring of 628 source point coordinates
will be placed in a circle with the radius specified in cell C2. The charge of each point
must be 1/628th of the total charge of the ring. This can be done simply by dividing the
charge by 628 and entering that constant number repeatedly. For the ellipse refer to the
major and minor axis for the X and Y coordinate formulas respectively.
1.
Place a uniformly distributed ring of charge with a radius of 0.25 meters in
the XY plane centered at the origin. The total charge on the ring is 35.0 nc.
What is the magnitude and direction electric field at each of the following
locations:
A. (0,0,0.5) m
B. (0,0,-0.5) m
C. (0,0,1.0) m
D. (5,0,0) m
E. (0,5,0) m
F. (0,0,5) m
G. (0,0,0) m
2. Replace the ring of charge with an ellipse with a major axis equal to 0.05 m
and a minor axis of 0.025 m. Repeat the same data points A-G from above.
3. Now replace the ring of charge with a single point charge of 35 nc at the
origin. Repeat data points A-F.
4. What must be the radius of a circular charge distribution of 0.45 c in order to
have a resulting electric field of 1000. N/C at (0,0,0.75)? Use goal seek.
5. A ring shaped conductor with radius a = 2.50 cm has total positive charge Q =
0.125 nC. At what distance along the central axis of the ring does the electric
field vector have a magnitude of 7.0 N/C. Use goal seek.