I need help with 5 Analysis section question c) and I need help with Comments on the distribution of the lines section with questions 1-5.

Lab: Equipotential Lines 1 Goal . To map the equipotential lines of a pair of charges on an electric board. . To draw the electric field lines. 2 Material . Electric potential mapping set (mounting board. conducting paper, special graph paper) . Battery . Connecting wires . Voltmeter Circuit A Circuit B V ++ +t T3 Theory 3.1 Electric field line Path along which a free electric charge would move inside a space region occupied by an electric field. 3.2 Equipotential line Line on which all points have the same value of elec- tric potential. Since these points have no potential difference between them, charges would not move from one point to the other. But, they would move between two points on two different equipotential lines. A positive charge would move from the higher to the lower potential and a negative charge would move in opposite direction. 3.3 Configurations A configuration of charges (+ and -) creates in the surrounding space an electric field with a particular geometry. Michael Faraday introduced the concept of electric field lines to visualize this field character- istics. The electric field lines are always perpendic ular to equipotential lines (or surfaces). They are emitted by "+" charges and go into "-" charges. Also, they are directed from the higher versus the lower values of potential.4 Manipulations 1. Draw the profile of electrodes (charges) on a paper (given by the instructor). 2. Connect a 6 V battery to the two posts on the mapping board, as shown. 3. Connect the negative terminal of a voltmeter to the negative post on the mounting board (negative side of battery). 4. Connect one end of a wire to the positive ter- minal of the voltmeter and use the other end of the wire to measure the electric potential point on the mapping board.5. Measure the potential of the negative (Vo) and Circuit A: positive (V6) posts. VO = 0.13 V6 = 3.87 6. Define potential steps as: AV = 16-V Sample calculation: 6 AV =(V6-VO)/6 = (3.87-0.13)/6 = 0.623 7. Find our values of interest: V1 = V0+1 AV = 0.13 + 0.623 = 0.75v Vi = Vo+ 1AV V2 = Vo + 2AV V3 = Vo + 3AV VA = Vo + 4AV V2 = 1.37v Va = Vo + 5AV V3 = 1.99v V4 = 2.61v 8. For Vi, find a few points (5 to 7) on the board V5 = 3.23v with the corresponding potential value. Make sure the points are well spread. Show their lo- cations on the map paper and draw a smooth Circuit B: line passing close by these points. This is the VO = 0.74 V6 = 4.91 AV= 0.695 equipotential line for Vi. V1 = 1.93v 9. Repeat the previous step with the other po- V2 = 2.13v tential values from step 7. You have to draw V3 = 2.83v at least 5 equipotential lines. V4 =3.52v V5 = 4.22v 10. Repeat the same procedure with the linear electrodes on the other side of mapping board.5 Analysis a) . Draw the set of equipotential lines (in blue). b) . From the set of equipotential lines, sketch the system of electric field (in red). c) . Calculate the magnitude (or the strength) of the electric field in four points selected in dif- ferent regions of each of the two maps by using the relation with potential change. Show these points on the map and the corresponding cal- culations for electric field. Comment on the distribution of these lines: 1. Are they straight or curved? 2. What is their direction? 3. Are equally dense or not? 4. Where is the magnitude of electric field big- ger? And smaller? 5. Compare your experimental findings with the- oretical predictions.