Question

9. Suppose an external force moves the +0.4C charge at constant speed from A to B following the path A to D to C to B. a. Along each segment of the path, find the work done by the electric force, FE, and compare to 3a above. Is the electric force conservative? Why? b.What is the change in electric potential energy, AU, along each segment of the path? What is the total ΔUE? c. What is the change in the electric potential AV along each segment of the path? d. The segments of the path that had no change in electric potential, AV-0, are called equipotential lines ° Identify the equipotential lines along this path from A to B. How are the equipotential lines oriented relative to the electric field lines? Why does this make sense? * Identify at least one other equipotential line on the graph. e. Suppose a-0.8C charge were moved instead of +0.4C charge. What are ΔUE and AV for the system when the negative charge moves along this path from A to B? f. If the -0.8C charge moved from A to C along the diagonal, what is the change in the electric potential energy? Why? E- 100 N/C 4

Answer 1

Work done along AD W_AD = FE times x = q E x = 0.4 times 100(-k) middot 4i W_DC = 04 times 100(-k) 2(k) = -80 Joule W_DC = 0.4 times 100 (-k)(-4i) = 0 Joule change in potential energy is negative of work done. in moving a charge from one point to another \r\n Delta V_AD = 0, Delta V_DC = 80 Joules Delta V_CB = 0 charge in electric potential E_r = partial differential /partial differential r partial differential V = E_r dr Delta V = -E. Delta r Delta V_AD = 0, Delta V_DC = -100 (2) = -100 volts Delta V_CB = 0 Equipotential lines rightarrow AD rightarrow CB Equipotential lines are oriented Perpendicular to electric field lines. Because no work is done in moving is charge Perpendicular to the direction of electric field lines + x axis will also be an Equipotential line when negative charge of - 8c moves Delta U_AD = 0,