# City Colleges of Chicago Electric Field due to a Point Charge Lab Report

Description

Charges and Fields
Electric Field due to a Point Charge
Concept: the electric field due to a point charge is given by
⃑𝑬
⃑ = 𝑲𝒒𝟐 𝒖
̂𝒓
𝒓
An electric field can be visualized on paper by drawing lines of force, which give an
indication of both the size and the strength of the field. Lines of force are also called
field lines. Field lines start on positive charges and end on negative charges
Procedure
Go to the web site

Once you are at the site “charges and fields” Click “play”.
The simulation contain the following items
A positive charge particle of 1 nC = 10-9 C
A negative charge particle of 1 nC = 10-9 C
A sensor that shows the value of the Electric Field at any point in space in V/m
A distance measuring tape in cm.
A grid that shows the direction of the electric field.
I.
Measurement of magnitude and direction of the Electric field due to
a point charge
The electric field for a point charge is given by
⃑𝑬
⃑ = 𝑲𝒒𝟐 𝒖
̂𝒓
𝒓
——— (*)
Where the constant k is given by K = 8.99 x 109 Nm2/C2
For the simulation q = 10-9 C. The magnitude of the electric field is going to be measured at
different directions and different distance r from the point charge. Note that the sensor in
the simulation gives the value of E in Volt/meter (V/m). It can be shown that 1 V/m= 1 N/C.
Notice the scale of 1 m in the grid in the lower left corner.
𝑲𝒒
̂𝒓 as E1, actually is an
Let’s denote the value for E obtained by the equation ⃑𝑬 = 𝒓𝟐 𝒖
experimental value because you need to measure r. Denote the value obtained by the sensor
as E2. Then calculate the % difference using the formula below
Note:
Percent difference is practically the same as percent error, only instead of one
“true” value and one “experimental” value, you compare two experimental
values. The formula is:
——–( **)
Procedure
1. Measure the Electric Field of the point charge in a direction of 0°
Move the positive point charge to the center of the plane. Assume this position as the
origin.
Click in the boxes in the upper right side to activate the electric field direction,
voltage, values, grid.
Use the sensor (yellow circle) to measure the Electric field at different points along
the x axis. The sensor gives the value of the electric filed in V/m
Complete the table below
X (m)
Distance from
the positive test
charge
E2 using the
sensor V/m
|𝑬𝟏 |
% error from
from equation (*) equation (**)
in N/m
0.5
1
1.5
2.0
2.5
3.0
3.5
4.0
2. Measure the Electric Field of the point charge in a direction of 90° with respect to
+x direction
Y (m)
Measured
vertically from
the positive
charge
0.5
1
1.5
2.0
E2 using the
sensor (V/m)
|𝑬𝟏 |
% error from
from equation (*) equation (**)
in N/m
3. Measure the Electric Field of the point charge in a direction of 45° with respect to +
x direction.
Use the measuring tape to verify the value of r. At 45°, follow the diagonal of the
square grid
r (m)
E2 using the
sensor V/m
|𝑬𝟏 |
from equation
(*) in N/m
% error from
equation (**)
0.705
1.41
2.12
2.82
4. Measure the Electric Field of the point charge in a direction of 45° with the negative
x direction
Use the measuring tape to verify the value of r. At 45°, follow the diagonal of the
square grid
r (m)
0.705
1.41
2.12
2.82
E2 using the
sensor V/m
|𝑬𝟏 |
from equation
(*) in N/m
% error from
equation (**)
Analysis
Do Excel plots
Plot E2 vs distance x. You have to make two plots, one for the results of part 1
and one for part 2. The plots must be a scatter plot. Do not joint the points
with a curve. Below is an example of the Excel plot
E from the sensor (N/m) vs x (m). Data of part 1
45
40
35
30
25
20
15
10
5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Questions
1. Do you obtain the same values for the electric field at directions of 0° and 90° for
the same distance?
2. Do you obtain the same value of the electric field for symmetric points at a
direction of 45° with positive x and at a direction of 45° with negative x?
3. Verify that the magnitude of the electric filed must be the same at points at the
same distance from the charge. From your data from part 1 and 2 complete the
table below
Use data from part 1 and part 2 to complete the table
X (m)
E2 using the
sensor (V/m)
from part 1
Y (m)
0.5
0.5
1
1
1.5
1.5
2.0
2.0
4. Write conclusions
E2 using the
sensor (V/m)
From part 2
% error
difference
using E2 for
the X and E2
for the Y
direction
from equation
(**)
II.
Electric Field due to two point charges.
To find the electric filed of two point charges at a given point in space, apply the principle of
superposition.
⃑𝑬
⃑ 𝒕𝒐𝒕𝒂𝒍 = ⃑𝑬𝟏 + ⃑𝑬𝟐 ;
——(***)
At a given distance r; E total is given by
⃑𝑬
⃑ 𝒕𝒐𝒕𝒂𝒍 =
𝑲𝒒𝟏
𝒓𝟐
̂𝒓 +
𝒖
𝑲𝒒𝟐
𝒓𝟐
̂𝒓 ——–(***)
𝒖
In the simulation the numerical values of q1 equal q2 are equal, and keep in mind that q2 is
negative
Procedure:
Locate both charges positive and negative separated a distance of 4 m. Assume the origin is
located at the position of the positive charge and place the positive charge to the left of the
negative charge
Calculate the coulomb force for a distance of 4 m
|𝑭𝒄𝒐𝒖𝒍𝒐𝒎𝒃| =
𝑲𝒒𝟏𝒒𝟐
𝒓𝟐
=
_______________ N
Find the total electric field of the two point charges along the axis that connects the
charges. Remember the origin is located at the positive charge, and positive x is the
direction to the right of the positive charge. Denote q1, r1 for the positive charge and q2 and
r2 for the negative charge
Complete the table:
r1 (m)
1
2
3
5
6
7
-1
-2
E1 (N/C)
From
equation
(*)
Direction
of E1
+X or -X
.r2 (m)
E2 (N/C)
From
equation
(*)
Direction
of E2
+X or -X
E total
from
equation
***
(N/C)
Direction
+X or -X
Complete the table using the sensor
.r1 (m)
Etotal using the
sensor in V/m
Direction +X or
-X
Etotal from the
previous table
(N/C)
% error
difference from
equation (**)
1
2
3
5
6
7
-1
-2
———(**)
Question
1. The % error difference increase, decrease or is random as function of
distance r.
2. Show that 1 V/m is equal to 1 N/C. Use the concept that 1 V = 1 Joule/C
3. Conclusions.
Series and Parallel circuits Lab
Name:
Series circuit
+

Parallel
+

Arrange bulbs in series and parallel circuits. Include pictures of each set-up.
In which set-up are the bulbs brighter?
Try is for both series and parallel, if one bulb is removed will the other go out?
Extra credit: If you have two batteries, can the batteries be arranged in series and parallel.
Which is brighter? Explain why this is.