ELECTRIC POTENTIAL & FIELD MAPS PURPOSE : To map the electric potential of different shapes of charged conductors, and compute the corresponding electric fields from these maps. APPARATUS: DC power supply, multimeter, three field map boards INTRODUCTION: The electric field E (a vector) at a given point in space is defined as the force in N that would be exerted on charge of +1 C. Calculating the force F exerted on a point charge of arbitrary magnitude q in an electric field E is very simple
F = q E
So E can be interpreted as the force per unit charge and its MKS units are N/C. Notice, the sign of q matters a lot! The force on a positive point charge will point in the same direction of E , while the force on a negative charge will point opposite E . Similarly, the electric potential V (a scalar) at a given point in space is defined as the electric potential energy difference, in units of J for a +1 C charge, with respect to a reference point. Calculating U of an arbitrary charge q in an electric potential V is very similar to calculating the force
U = qV So V can be interpreted as the potential energy per unit charge and its MKS units are J/C. This unit is referred to as a volt and is abbreviated with V (be careful not confuse the unit with the potential ). Just like the potential energy U, the value of V depends on the reference point you pick. Generally, we are uninterested in the value of V itself, just how it changes from point to point. Often, a reference point is selected where the value of V is zero. Charges are not only affected by electric fields and potentials, but they also generate them. Recall Coulomb’s law from the last lab, which describes the force between two different point charges. Let q 1 be located at the origin, while q 2 is a distance r away from q . Using magnitudes only for simplicity, we see that the force exerted on q by q is 1 2 1 2 2 |F k|q q |)/r = |q |(k|q |/r ) = |q || E → | E |/ r2 21| = ( 1 2 2 1 2 1| 1| = k|q 1
So E is the electric field generated by q 1 1 and its magnitude is given by the above expression. E p oints radially a way from q if q towards q 1 1 1 is positive, and radially 1 if q 1 is negative. We follow a similar logic using the potential energy between q 1 and q 2 to get V ; the electric potential g enerated by q 1 1
U = kq q /r = q (kq /r) = q V → V = (kq /r) 21 1 2 2 1 2 1 1 1
V 1 has no direction because it’s a scalar, but it is positive or negative everywhere depending on the sign of q . Another thing worth noting is that the reference point for the 1 above calculation is at infinity, a common and convenient choice in electrostatics.
Using the definition of potential energy, one can show that the electric potential and field are related by
V = ∫ E • ds
This integral definition shows us that we integrate E over a path to yield V in the same way we integrate F over a path to yield W, the work done by force F . But for the purposes of this lab it’s better to use a differential definition
E = ▽ V or (E , E , E ) = (∂V/∂x, ∂V/∂y, ∂V/∂z) x y z
This above expression suggests another convenient way to express the MKS units of the electric field, V/m. Note; proof of these expressions is not important for this lab, regard them as given until your lecture catches up! Let's list some properties of conductors in electrostatic equilibrium since we will be using (and have used) conductors in the lab: • All points on and inside a conductor have the same V, i.e., conductors are equipotentials (defined below). • E directly outside a conductor's surface is perpendicular to the surface at every point, and zero everywhere inside the conductor. • Any net charge on a conductor always lies on the surface. Electric Field Lines : Electric field lines are a way of visualizing E near charged objects. The field lines follow the direction of E . The lines are drawn such that the number of lines per unit area is proportional to the magnitude of E . Equipotential Surface : A surface of constant V. Often simplified to equipotential. Equipotential Lines : Equipotential lines are a way of visualizing equipotentials on paper. If the charge q were able to move along an equipotential line very slowly , there would be no change in the electrical potential energy of q (recall that W = ▵U, since U = qV, W = q▵V = 0). Thus, E would do no work on the charge! From the definitions of equipotentials, E , and V, some properties of the lines follow: • Electric field lines are always perpendicular to equipotential lines. • Electric field lines point from higher equipotential lines to lower ones. • Equipotential lines describe a path and/or surface along which a charge can move with no external work being done. Let’s consider an example, see Figure 1 on the next page, which shows a charge q at the origin with it’s electric field lines (solid) and equipotentials (dashed) already drawn. Let 9 |q| = 1.0 nC, the equipotential lines be separated by a distance of 3 cm. Use k = 9 x 10 N 2 2 m /C for Coulomb’s constant for simplicity.
FIGURE 1
Let’s place another charge q 0 = 3 nC on the positive xaxis, on the outermost equipotential line drawn in the figure. (1) Calculate the E due to q (magnitude and direction) at the location of q . q 0 2 2 ANSWER: | E | = k|q|/(15 cm) = 4 x 10 N/C, pointing to the right q (2) Is the force between the charges attractive? Calculate F , the force on q 0 0 by q (magnitude and direction) using E . q ANSWER: Yes it is attractive, the charges have opposite signs! 6 | F0 | = |q | | Eq | = 1.2 x 10 N, pointing to the left. 0 (3) Calculate the electric potential due to q at the location of q . 0 2 ANSWER: V = kq/(15 cm) = +1.8 x 10 V q (4) Calculate the electric potential energy U, between the two charges. 7 ANSWER: U = q V = 5.6 x 10 J 0 (5) Electric potentials and fields obey the principle of superposition. What is the electric potential due to both charges at 6 cm on the positive yaxis? 2 2½ 1 ANSWER: V = V + V = kq /{[(6 cm) + (15 cm) ] } + kq/(6 cm) = 9.4 x 10 V tot 0 q 0 (6) How much work would it take for an outside force to move q 0 to the innermost equipotential line? Does your answer change if the path taken by q 0 isn’t a straight line? 7 ANSWER: W = q ▵V = q (V V ) = kqq (1/(3 cm) 1/(15 cm)) = 7.2 x 10 J 0 0 q,in q,out 0 W only depends on the initial and final distance between q and q 0
PROCEDURE: For each configuration you will map V, locate and plot four equipotential lines (4, 8, 12, and 16 all in Volts), and sketch E . See the next section on how to draw 3D plots using Origin Pro. In all cases be careful to take sufficient data at regular spatial intervals in both horizontal and vertical directions to accurately define the shape of the equipotential lines, and thus their corresponding E . Since each configuration is symmetric, it is sufficient to map V for one half of each configuration, and sketch the other. The three conductor configurations are:
1. two points 2. parallel plates 3. parallel plates with a circular conductor in the middle
In practice, E is difficult to measure directly. So instead, we measure V with a multimeter, relative to the negative conductor, which is held at a constant potential using the power supply and therefore acts as our reference point. Later, components of E can be calculated by taking the derivative. Since measurements don’t give us a function to explicitly differentiate, we approximate components of E like E V/▵x and E x ≈ ▵ y ≈ ▵V/▵y. Figure 3 shows one possible experimental arrangement for simulating the electric field between two point charges. Connect your power supply (set to 20 V at most) to the two pointlike conductors on the board using the positive and negative leads (red and black respectively).
FIGURE 3: The voltmeter above represents the multimeter measuring voltage.
The set up in the lab is slightly different from figure 3. First, the power supply only goes as high as 18V. Second, we can’t connect the multimeter directly to the power supply. Instead, an equivalent set up is to fix the multimeter’s negative lead (again, black) to the negatively charged conductor, and set the multimeter to read DC voltage. Now the positive lead on the multimeter may be used to measure V by gently touching it at any point on the board.
Do not write on, scratch, or gouge the conducting paper on the board. Replacement is very costly.
To begin, note the voltage that you set the power supply to. Then, try taking some preliminary measurements. Try touching the positive lead of the multimeter to the negatively charged conductor, what does it read? What should it read? Then, see if you can locate an equipotential. For the two point configuration, measure between the points as well as around them, the equipotential lines you get should be that of a dipole. For the parallel plates, measure potential in between and outside of the plates. For the plates with a circular conductor, make similar measurements but also take measurements inside the ring as well. APPENDIX: MAKING A 3D GRAPH USING ORIGINPRO All data are to be entered in an OriginPro matrix worksheet. Do not include the matrix worksheet in your write up. The 3D plotting routine in ORIGINPRO will plot the value of a cell on the zaxis and the location of the cell as the x and y location. For example, if cell C(11,3) has the value 17, then ORIGINPRO will plot x = 11, y = 3, and z = 17. In this experiment the value of Z represents the potential. To make a 3D plot, first you have to create a 2 dimensional matrix. These two dimensions represent x and y axes. The matrix entries represent the values of z (i.e. measured potential). To create one, go to “File”, from the drop down menu select “New” and then “matrix.” A window pops up where you can select the number of y rows and x columns. You can also select the range of values of x and y. In the box for “Numbers of Matrix Objects in Sheet”, enter 5. After selecting these values, click “OK”. A new window pops up with an empty matrix. Now you can enter the values of potential corresponding to each (x,y). After doing so, click on “Plot”, from the drop down menu select “3D surface” and “color fill surface”. You can change the orientation of the plot by rotating it, and also the size of each axis by zooming the appropriate axes. To display equipotential lines, select another matrix object and populate all of its elements with, for example, the potential value 4(V). Click the 3D plot you made, then select Graph menu and select plot setup. Add the second matrix object to the plot. You can then display both the potential data surface and the surface for 4 V. The intersection of the two surfaces would be the equipotential line. To find other equipotential lines, make other matrix objects populated with other values of potential. Connect points on your equipotentials with smooth lines to produce equipotential lines. Label the potentials. The E field lines should be everywhere perpendicular to the equipotential lines, from the defining relation between E and V. The direction of E is toward decreasing potential V. Sketch the E field lines on your plot.
ELECTRIC POTENTIALS & FIELD MAPS
Names: Section: Date: You may do the maps in any order. Refer to the procedure above. Your answers should be brief, neat, wellwritten sentences. Include for each configuration , an Origin printout showing 3D graph of V vs. x and y, the equipotential lines (at 4, 8, 12 and 16 V), and sketches of the electric field. Additionally, pick three points and compute the magnitude and direction of E at each. Indicate the locations of the conductors and which leads (positive or negative) they were connected to. Configuration One: TWO OPPOSITE POINT CHARGES a) Where is E most nearly uniform?
b) Describe the shape of the central equipotential line near the midpoint of the conductors.
c) What is the equipotential shape close to the "point" electrodes? Configuration Two: TWO PARALLEL PLATES a) Describe the shape of E inbetween the plates.
b) Does E extend beyond the edges of the plates?
Configuration Three: HOLLOW CIRCULAR CONDUCTOR a) Is there a net charge on the circular conductor? Give an argument based on the shape of E and the equipotential lines. See the introduction for properties of conductors in electrostatic equilibrium to help craft your argument.
b) What is V on the conductor ring? What is V inside the ring?
c) What should E be inside the conducting circle? Is the inside of the circle “shielded” from E as expected? Why or why not?
