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Lab #3 – Electric field and electric potential Honghao Ma Partner: Yang Liu Date: Feb 10, 2021 Abstract: In this lab, we observe the relationship between ε0 and charge through the experiment of Electric field of one charge, and confirm its value, and then observe the relationship between Electric field and potential through experiment. Experimental setup and measurement: In this lab, we mainly used the website, which is the “Charges and fields” simulation (https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_en. html). And measure the electric field, electric potential, distance and Potential values according to this website. Experimental data: Part2:
The value of the electric field = 38.2 V/m Q = +1 nC = 1 * 10 -9 C <----------------------------------------------------------------------------------------------------------------> Part3: Case No. Electric field (V/m) Distance(m) 1 38.2 0.5 2 14.4 0.8 3 8.93 1.0 4 6.36 1.2 5 3.99 1.5 6 2.28 2
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Electric Field (V/m) Electric potential (V) 39.3 18.5 9.05 9.1 4.02 6.0 2.23 4.5
<----------------------------------------------------------------------------------------------------------------> Part4:
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6 more potential values: <----------------------------------------------------------------------------------------------------------------> Data analysis: Part2: 1. The force directs to the negative charge. And the force increases with the electric field sensor closing to the negative charge. The force disappears when the electric field sensor and negative charge coincide. 2. The force direction points away from the positive charge. And the force increases with the electric field sensor closing to the positive charge. The force disappears when the electric field sensor and positive charge coincide. 3. E = (Q/4π*ε0)*X, ε0= (Q/4π*E) *X, where X = 1/ r 2 and Q = 1* 10 -9 Distance(m) 1.1 1.3 1.5 1.7 1.9 2 Potential values(V) 160.7 145.8 133.5 123.0 114.0 109.6
Slope = 9.6103 = Q / (4π*ε0) = 1*10 -9 / (4π*ε0) So ε0 = 8.28*10 -12 ± 0.31 F/m Expected value ε0 = 8.8541878128(13)×10 −12 F ⋅ m −1 <----------------------------------------------------------------------------------------------------------------> Case No. Electric field (V/m) Distance(m) X = 1/ r 2 1 38.2 0.5 4 2 14.4 0.8 1.563 3 8.93 1.0 1 4 6.36 1.2 0.694 5 3.99 1.5 0.444 6 2.28 2 0.25
Part3: 1. When moving the ‘Equipotential meter’ along these lines, the value of each point on these lines are equal. 2. When moving the electric field sensor along these lines, the value of the electric field is the same with the equipotential line and its orientation is perpendicular to the equipotential line. 3. When the distance from the charge decreases, both the electric field and electric potential are increasing. The differences are that the field strength has a direction and is a vector while the electric potential is relatively defined. 4. We found that the electric potential has a certain functional relationship with the electric field. Electric potential = electric field * d, where d is the distance between two points along the electric field line. 5. Electric Field (V/m) Electric potential (V) EP / EF 39.3 18.5 0.467 9.05 9.1 1.006 4.02 6.0 1.493 2.23 4.5 2.018
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6. We think that there is a point where the electric field is zero. And the point is in the middle of these two positive charges. One meter away from each other. 7. From the picture, we got the point where the electric field is zero. The point is in the middle of the distance between two charges. The analog from mechanics that we come up with is two objects collide with each other. Because the forces are mutual, the forces received by these two objects are equal but opposite. So we guess that there is 0 at some point between these two objects. When moving it away from these two charges, the equipotential line wrapps the two charges. Gradually, it changed from an ellipse with two charges as the focal point to a circle with the midpoint of the two charges as the center. 8. Sketch the electric field and equipotential lines for this case of two positive charges
9. After changing one of the positive charges with a negative charge, there is no point where electric field is zero.
10.Sketch the electric field and equipotential lines for this case of one positive charge and one negative charge.
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11.Sketch the electric field and equipotential lines for this case of a dipole. <----------------------------------------------------------------------------------------------------------------> Part4: 1. The difference between 1.0m and 0.5m potential value = |169.5 - 236.6| = 67.1V 2. Potential value on respection point = 236.6V V = [30 / ( 2π*ε0 )]* X
Potential difference(V) vs. X = ln(r0/r) Slope = -86.658 = 30 / ( 2π*ε0 ) So ε0 = 30 / 2π* (-86.658 ) = =0.0551 <----------------------------------------------------------------------------------------------------------------> Distances(m) X = ln(r 0 /r) Potential values(V) Potential difference(V) 1.1 -0.79 160.7 75.9 1.3 -0.96 145.8 90.8 1.5 -1.10 133.5 103.1 1.7 -1.22 123.0 113.6 1.9 -1.34 114.0 122.6 2.0 -1.37 109.6 127
Part5: Conclusion: By measuring and applying equation for the magnitude of the electric field We can confirm the value of the vacuum dielectric constant ε0 = 8.8541878128(13)×10−12 F ⋅ m−1. And by observing the experimental graphs of electric field and electric potential, we can find that there is a certain functional relationship between them.
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