Potentials Worksheet The sketch below shows a cross section of equipotential sur
ID: 1582676 • Letter: P
Question
Potentials Worksheet The sketch below shows a cross section of equipotential surfaces between two charged conductors. 50 V -40 V +20 V +40 V -30 V 20 V 1. Draw 5 electric field lines, starting on the plate on the right, connecting the two conductors. Include the direction. 2. At which of the labcled points will the electric ficld have the greatest magnitude? A B CD E F G H 3. At which of the labeled points will an electron have the greatest potential energy A B C D E F G H I 4. What is the potential at point E? Vr= 5. What is the magnitude of the potential difference between points B and E? 6A-1.0 C charge is moved from A to E? How much work is done by the electric field? 7, A-1.0 C charge is moved from B to D to C. How much work is done by the electric field? WBDC =Explanation / Answer
(a)draw any five lines from conducting plate to point charge. As conducting plate is positively charged so electric field lines will emerge out from it and point charge is negatively charged so electric field lines will enter into it.
(2) point nearest to charge where absolute value of potential is highest - at point H
(3) electric potential energy of electron = -eV
Where -V is maximum I.e. at point H
(4)point E is on equipotential surface having potential -40 Volt so potential at point E = -40 Volt
(5) VB - VE = 10 - (-40) = 50 Volt
(6) W = qVf-qVi = -10-6(-40-30) = -7×10-5 joule
(7) potential of initial point and final point is same . So change in potential is zero and hence work done is also zero.
(8) positive charge will have a tendency to move toward negative charge or less potential so when a positive charge is placed at point F it will move toward H (d)
(9)E = -V/r = 40/0.4 = 400 N/C leftward
(10) W = qV
50×10-6 = q(-50-(-10))= -40q
q = 1.25×10-6 C
(11) Vg-VB = -60Volt
Work done on charge = 3×10-6×60 = 180×10-6 joule
1/2mv2 = 180×10-6
V = 10.95m/sec
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