Electric Field and Potential Lab yip

1 (1 point) Title of the Experiment: Electric Field and Electric Potential Student’s name: Puurich Yip Section SLN:13710 TA’s Name: Sang-Eon Bak, Francesco Setti Week of the experiment: LAB 2

2 OBJECTIVE ( 3 points ): The purpose of this lab is to discover electric field strengths and potential in an electric field. This lab will also help us determine the electric field strength of 2 charged plates. EXPERIMENTAL DATA ( 6 points ) & 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! PART 1A: Electric field of the point charge distributions: a. Note: use the same values of q and r for part 1A(a) through part 1A(d). • 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 = -1nC

4 [See Module 3 on Canvas for q and r values] - Explain your observations qualitatively and qualitatively: What happens to the electric field when the charge is doubled while the distance is kept constant? When the charge is doubled and the distance is kept constant, the electric field strength also doubled. c. The labeled screen capture of the charge distribution at different positions with a constant charge. - The value of the charge q = -1nC Distance r (m) E meas (V) E cal (V) r = 1.5 -5.97 -6 2r = 3 -2.99 -3 [See Module 3 on Canvas for q and r values] - Explain your observations qualitatively and qualitatively: What happens to the electric field when the distance is doubled while the charge is kept constant? The electric field weakens by 50 percent when the distance is doubled. d. The labeled screen capture of the charge distribution (with two charges) - superposition principle.

5 r 1 = r 2 = r = 1.5m) ; q 1 = q 2 = q = -1nC [See Module 3 on Canvas for q and r values] Where E 1x,meas 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 E 2y,meas . Calculate the theoretical values for E x,net , E y,net , E net , and θ for the charge distribution using the vector superposition principle: E x,net = E 1x,calc + E 2x,calc = -6V E 1x,meas (V) E 1y,meas (V) E 2x,meas (V) E 2y,meas (V) E net,meas (V)  meas -5.98 0 0 -17.76 -11.78 45 degrees

7 - How does the potential vary with distance from the source? Is the graph of V vs. r linear? The potential weakens exponentially as the distance from the source increases. The graph is not linear. • Insert the graph of V vs. 1/r Slope: m = 7.2m Compute the quantity kq = 7.2m % error between the slope m and kq (taking kq as the actual value) = 0 • V vs. r at different equipotential surfaces with negative charge : r (m) 0.5 1.0 1.5 2.0 2.5 V horizontal (V) -17.98 -9.0 -6 -4.5 -3.59 V vertical (V) -18.0 -8.99 -6.02 -4.48 -3.61 V ave (V) -17.99 -8.995 -6.01 -4.49 -3.6

8 • Insert the graph of V vs. r - How does the potential vary with distance from the source? Is the graph of V vs. r linear? The potential has a direct relationship with the distance from the source. The graph is not linear, its exponential • Insert the graph of V vs. 1/r -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 0 0.5 1 1.5 2 2.5 3 Vave (V) R (m) v vs r 0 0.5 1 1.5 2 2.5 -20 -15 -10 -5 0 Vave(V) 1/r(m) v vs 1/r

10 - Is there any symmetry? The fields are symmetrical, and opposite because of the opposite charges - How are the electric field vectors oriented relative to the equipotential lines? The field vectors are oriented in a circular pattern, pulled more towards the negative charge. PART 2: Electric field and electric potential between parallel plates A- Wand Charge # Q 1 = 105 [See Module 3 on Canvas for Q 1 value] - The value of the charge on the ball as it appears in the info box: 5.05*10^-9 C - Actual charge on pith ball (negative of value from info box) 𝒒 ? = -5.05*10^-9 C - The value of the mass of the ball as it appears in the info box: 0.05 grams - The angle θ 1,max between the string and the vertical line: 6 degrees - Separation distance d between the parallel plates using ruler: 8cm - Free body diagram the ball

11 - Apply Newton’s 2 nd Law the charged pith ball in equilibrium position to state the two equations of the equilibrium state Tcos(6)=mg qE=Tsin(6) - Using the equations and the value of the charge and mass, calculate E 1,max ( add the Unit ) T=5.11x10^-4 N E=282.7 V - Calculate V 1,max using the values of E 1,max and d ( add the Unit ) - E1max=Tsin(6)/q = 5.11x10^-4 N(sin(6))/ 5.05*10^-9 C = 28,273.5N/C - V1max= E1max d = 28,273.5N/C(0.08m)=2261.9 V With θ 1 = 6 degrees [See Module 3 on Canvas for θ 1 value] Use the equations to calculate - The electric force F 1 = 5.11x10^-4*sin6=1.42x10^-4 - Electric field E 1= 282.7 V - Electric Potential between the plates V 1 = 2261.9 V Hint: Use the fact that F = q*E and |E| = |  V/  x| B- Wand Charge # Q 2 = 90 [NOT SURE HOW TO DO THIS PART, KET SIMULATION DOESN’T WORK AND THE VIDEO DOESN ’ T SHOW VALUES FOR WAND CHARGE 90] - The value of the charge on the ball as it appears in the info box: …………………… (unit) - Actual charge on pith ball (negative of value from info box) 𝒒 ? = …………………… ( unit) - The angle θ 2,max between the string and the vertical line: ………………………………. - Using the equations and the value of the charge and mass, calculate E 2,max ( add the Unit ) - Calculate V 2,max using the values of E 2,max and d ( add the Unit ) - % difference (not % error) between V 1,max and V 2,max With θ 2 = ……………………………… [See Module 3 on Canvas for θ 2 value] Use the equations to calculate