OBJECTIVE (3points):
What are the physics concepts/theory/law to be investigated in this experiment? It is one or two sentences in your own words.
EXPERIMENTAL DATA (6points) &DATA ANALYSIS & RESULTS (10 points)
Obtain experimental data that will be used for further calculations from the graphs or tables
Be sure to show your calculations and to include related equations and diagrams!
Objectives
• To explore electric fields around different configurations of charges and map their
Electric fields and equipotential lines.
• To explore the behavior of a charged pith ball suspended in a uniform electric field of
Parallelplates.
• To experimentally determine the strength of the electric field and potential difference between two oppositely charged parallel plates.
• Determine the magnitude and direction of the force on a charged particle in an electric field.
Theory of the experiment
Law of the experiment
The law being investigated in this experiment is electrostatics law which states that “like charges repel while unlike charges attract”
PART 1A: Electric field of the point charge distributions:
a. The labeled screen capture of the charge distribution with the four sensors (two on vertical and two horizontal axes).
• The value of the charge q = …………4.5………. (nC)
• The distance of each sensor from the point charge r = ……10……………. (m)
Position +x (right) -x (left) +y (upward) -y (downward) Eave
Emeas (N/C) 0.65 0.45 0.54 0.35 0.4975
Ecal (N/C)
0.45 0.30 0.41 0.33 0.3725
Where Emeas are the values appear on the screen and Ecal are the values calculated using equation (3).
Show the calculation of one Ecalvalue here:
% error of the two average values (assuming the calculated as the accepted value) = …12.5%………………..
• Explain your observations qualitatively:
What happens to the magnitude of the electric filed as you place the sensors closer or farther? Talk about directions of the E. field at the different locations of the sensors.
The strength of an electric filed is affected by source charge, which is inversely related to the square of the distance from the source. The strength of the electric is dependent to location hence decreases as the distance from the source to location increases
b. The labeled screen capture of the charge distribution while doubling the charge at constant distance.
– The distance of the sensor from the point charge r = …10……….. m
Charge q Emeas (N/C) Ecal (N/C)
1.0 nC 0.21 0.10
2.0 nC 0.24 0.20
– Explain your observations qualitatively and qualitatively:
Both the observed and calculated values as distance is varied but with little variation between the two values hence high level of accuracy.
What happens to the electric field when the charge is doubled while the distance is kept constant?
Doubling the charge increases the electric field since it is directly proportional to the charge
c. The labeled screen capture of the charge distribution at different positions with a constant charge.
– The value of the charge q = …………………. (unit)
distance r (m) Emeas (N/C) Ecal (N/C)
0.5 0.021 0.04
1.0 0.013 0.032
– Explain your observations qualitatively and qualitatively:
There is slight change in the value of electric field as the charge is doubled. This result to small variation in the observed values and the calculated values of electric field.
What happens to the electric field when the distance is doubled while the charge is kept constant?
Increasing the distance reduces the electric field
d. The labeled screen capture of the charge distribution (with two charges)- superposition principle.
r1 = r2 = 1.0 m. ; q1 = q2 = 1.0 nC
E1x E1y E2x E2y Enet,meas meas
0.45 0.33 0.30 0.35 1.43 1.78
Where E1x is the x-component of the electric field at the sensor due to the presence of the first charge (to the left) only. Same thing for E2y.
Ex,net = E1x + E2x = 0.75
Ey,net = E1y + E2y = 0.68
Enet = 1.27 ; % error with Enet,meas = 11.18%
cal = 2.05 % error with meas =.15.17%
– Explain your observations qualitatively and qualitatively:
What happens to the electric field when you add one more charge (third) to this configuration?
Compare it with the presence of only one charge above the sensor.
PART 1B: Electric potentials and equipotential lines of point charge:
• V vs. r at different equipotential surfaces with positive charge:
r 0.5 1.0 1.5 2.0 2.5
Vhorizontal 12.0 9.0 7.0 4.5 3.0
Vvertical 11.0 7.0 4.5 3.0 1.5
Vave 11.5 8.0 5.8 3.8 2.3
– Insert the two graphs of V vs. r and V vs. 1/r
• V vs. r at different equipotential surfaces with negative charge:
r 0.5 1.0 1.5 2.0 2.5
Vhorizontal 10.0 7.0 4.5 3.0 1.5
Vvertical 9.0 6.0 3.0 2.0 1.0
Vave 9.5 6.5 3.8 2.5 1.3
– Insert the two graphs of V vs. r and V vs. 1/r
PART 1C: Electric field, potential and equipotential lines of electric dipole:
– Insert the graph of equipotential lines around each charge with the magnitude of the potentials 1.5 V; 3.0 V; 4.5 V; 7.0 V and 10.0 V
– Is there any symmetry?
Yes
– How are the electrical field vectors oriented relative to the equipotential lines?
PART 2: Electric field and electric potential between parallel plates
A- W/ Charge # = 120
– The value of the charge on the ball as it appears in the info box: 6.168e-9 C
– The value of the mass of the ball as it appears in the info box: 0.050 grams
– The angle θ between the string and the vertical line: 8o
– Separation distance d between the parallel plates using ruler: 2.0 m
– Free body diagram the ball
– Apply Newton’s 2nd Law the charged pith ball in equilibrium position to state the two equations of the equilibrium state
F = mq1 q2
4ΏE0 r2
E = mq
4ΏE0 r2
– Using the equations and the value of the charge and mass, calculate Emax (add the Unit)
EMax = 0.05 × 10-2 × 6.168×10-10
4 ×Ώ× 8.5×10-12× 4
= 7.218×10-4 N/C
– Calculate (Vmax)1 using the values of Emax and d(add the Unit)
=VMax =EMax × d
=7.218×10-4 ×4
=2.8872 ×10-3 V
With θ = 8o
Use the equations to calculate
– The electric force F1
=F1 = k|q|
r2
= 8.99×109×6.168 ×10-9
4
= 13.86258N
– Electric field E1
E1 = F
q
= 13.86258
6.168 ×10-9
= 2.247 ×109 N/C
– Electric Potential between the plates V1.
Use the fact that F = q*E and |E| = |V/x|
V1 = E × change r
=2.247× 109 × 4
=8.90 × 109 V
B- W/ Charge # = 100
– The value of the charge on the ball as it appears in the info box:4.692e-9 C
The angle θ between the string and the vertical line: 8o
– Using the equations and the value of the charge and mass, calculate Emax (add the Unit)
EMax = mq
4ΏE0 r2
4.692 ×10-9 × 0.05 × 10-3
4 ×Ώ ×8.5 ×10-12×4
= 1.725 ×10-3 N/C
– Calculate Vmax using the values of Emax and d (add the Unit)
=VMax =EMax × d
= 1.725 × 10-3 ×4
= 6.9 ×10-3V
– % difference between (Vmax)1 and (Vmax)2
= VMax2 – VMax1 = 6.9 × 10-3 – 2.8872 ×10-3
= 4.0128 × 10-3 V
V2 = 6.9 ×10-3
2.8872 ×10-3
V1
=2.389 V
With θ = 8o
Use the equations to calculate
– The electric force F2
– Electric field E2
– Electric Potential between the plates V2.
– V2/V1 = 2.389V
– % error of V2/V1 compared to 1.3 = …2.389V………….
Screen Shots from KET Lab
DISCUSSION & CONCLUSION (10 points):
This is the most important part of the lab report. It is where you describe whether your results support the physics principal being investigated in the lab. Begin the discussion with the purpose of the experiment. Briefly explain the theory concept that was tested. Then state only the key results (with uncertainty, percentage or difference errors, and units) quantitatively with numerical values; do not provide intermediate quantities. Discuss the relationship between your raw measurements and your results; the relationship between quantities in the graph; relationship between the independent and dependent variables. All questions from the lab manual should be answered in the narrative form.
An array of electric charges wields forces on one another through turbulence in their contiguous space known as electrical fields. From Coulomb’s Law, the energy which an electrical field puts on a charge. Through the Theory, it is likely to compute the electric field of different charge distributions through using Coulomb’s Law and the supporting principle. In practice, nevertheless, it is slightly complex to figure it because of the vector nature of the fields or forces.
The electric potential is refers as the prospective, stimulating power U divided by the charge q: V = U/q. Thus the electrical potential has scalar magnitude with SI units called the volt (V), where 1 V = 1 joule/coulomb. If the electrical field is known, we can determine the electrostatic potential of any arbitrary point charge. So if you see the potential, you can calculate the area and vice versa. Due to the scalar nature of the electric potential, it is easier to work with. The scale of an electrical field in an area can be projected through measuring the probable difference ∆V over a displacement ∆d. Equipotential areas are areas at a continuous potential (or voltage). In three dimensions (3D), equipotential parts are often surfaces, providing a graphical 3D representation of the possible. In two sizes (2D), equipotential areas are often shape or planes called contours. Electric fields and equipotentials have the following relations: The electric field is perpendicular to the equipotential surfaces, the field points in the direction of decreasing potential and equipotential lines never cross each other and electrical field lines never cross each other. to the course of the electric field lines.