Physics 241 Lab: Electric Field Mappinghttp://bohr.physics.arizona.edu/~leone/ua/ua_spring_2010/phys241lab.htmlName:____________________________Section 1:1.1. One view of particle dynamics is to think about how a particle responds to the forces exerted onit (the vector force field). Another equally valid point of view is to may think of how a particleresponds to the potential energy landscape that surrounds it (the scalar energy field). An example ofhow particles move in one dimension is shown below. Imagine that some external force creates the“hilly” potential energy graph shown, and that four particles are placed at locations A through D. The particle at A is at a minimum of the potential energy curve so that if it is moved, theparticle will rise in potential energy. Therefore, the particle at A remains at rest. The particle at B sitsat a part of the potential energy curve that has a positive slope. If it moves to the left, it can lower itspotential energy. Therefore, the particle at B feels a force pushing it to the left. The particle at C sits ata precarious position. The slope of the potential energy curve at C is zero so that the particle feels noforce. However, any small perturbation in the particles position will cause it to “tumble” from itsunstable equilibrium. The particle at D has been placed where the slope of the potential energy curveis negative. If the particle moves to the right, it will lower its potential energy. Therefore, the particleat D feels a force pushing it to the right. Notice that the magnitude of the slope at B is greater than atD. Thus the particle at B does not have to move as far to lower its energy as the particle at D. Thiscorresponds to the particle at B feeling a stronger force than the particle at D.The big idea here is that given a graph of potential energy, you can find the corresponding forceat each location by realizing that Force equals the negative slope of the potential energy, € r F (x) = −dU(x)dxˆ x . If a potential energy is given as € U(x) =12k x − 2( )2 where € k = 3 [J/m2], what isthe force from this potential energy on a particle placed at x = 1 m? At which positions is the forcezero?Your calculations and answers in SI units:1.2. The previous technique may be applied to motion in more than one direction:In this situation, the particle can move in the x and y directions. The potential energy landscape showsa single unstable equilibrium (“mountaintop”). Dotted lines have been drawn to show where thepotential energy is constant. These are called equipotential lines. If this were a topographical map,these lines would represent constant height (called contours), and mountaineers maneuvering aroundthe mountain at a constant height would be “contouring”. The force on the particle “down themountain” is always perpendicular to the equipotential lines. Again the force will equal the negative slope of the potential energy graph, but how do youdescribe a two-dimensional slope? The answer is that the slope will be a two-dimensional vector (so isthe force!), which can be found using partial derivatives: € r F (x, y) =−∂U(x, y)∂xˆ x +−∂U(x, y)∂yˆ y .This equation makes intuitive sense. If you want the component of the force in the x-direction, findout how much the potential energy is changing in the x-direction using the partial x-derivative.If a potential energy is given as € U(x, y) = k 2x2+ xy + 3xy2( ) where € k = 1 [N/m], what is the force inthe x-direction from this potential energy on a particle placed at € x = 2 m and € y = 0 m? Your calculations and answer in SI units:1.3. When charge is placed on a conductor, the excess charge creates an electric (vector) fieldaround it. If a test charge q is place in this electric field, the test charge feels an electric force € r F = qr E .Another equally valid point of view is that the charged conductor creates an electric potentiallandscape (a scalar field):Just as the force was the slope of the potential energy graph, the electric field is the slope of the electricpotential graph: € r E (x, y) =−∂V (x,y)∂xˆ x +−∂V (x, y)∂yˆ y or € r E =−∂V∂xˆ x ,−∂V∂yˆ y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟. In a laboratory,one approximates derivatives with finite difference measurements: € r E (x, y) ≅−ΔV (x, y)Δxˆ x +−ΔV (x, y)Δyˆ y =−(V (x + Δx, y) −V (x,y)Δxˆ x +−(V (x, y + Δy) − V (x, y))Δyˆ y This means that you can approximate the electric field at a point in space by using a DMM to measurethe negative change in the voltage in each direction divided by the distance of your measurement.These ideas are easily extended to three dimensions: € r F (x, y,z) =−∂U(x, y,z)∂xˆ x +−∂U(x, y,z)∂yˆ y +−∂U(x, y,z)∂zˆ z c (mathematically identical)r E (x, y,z) =−∂V (x, y,z)∂xˆ x +−∂V (x, y,z)∂yˆ y +−∂V (x,y,z)∂zˆ z 1.4. In this lab, we will be working with two-dimensional electric vector fields (using conductingpaper), but we will keep are graphs of the electric field and equipotentials restricted to two-dimensionsby “looking at the mountain from above”. This “aerial view” will make it easier to make graphs, butdon’t forget the underlying principles!Section 2: Some people find the above picture useful for connecting the various electrostatic concepts together and even add to it as they learn more.2.1In the above schematic, a formula is given for the electric field. For what physical situation is this formula correct/useful? Your answer:For what other physical system(s) would the electric field be best found using a different formula or mathematical technique? Your answer:Many introductory students often only know one method/formula for finding E and get stumped on their exams!Section 3:3.1. In the following picture depicting equipotential lines (dashed lines of constant voltage), sketchthe corresponding electric field lines (draw the electric field lines with solid lines). Be sure to label thedirection of the electric field lines using arrows using the assumption that VOUT < VIN. Note that Voutgives the value of the voltage for the entire dotted line it touches (similarly for Vin). Thinktopographical maps! Draw on the picture below.In the following picture assume VOUT <
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