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Lecture Notes | Approved: 7 years ago | 913.5 kB | Comments: 0
...symmetry should simplify the calculations. Field Due... ...Use spheres as the gaussian surfaces When... ...When r a, the charge inside the... ...the charge inside the surface is Q... ...When r a, the charge inside the... ...the charge inside the surface is 0... ...of Charge, cont. The end view confirms... ...end view confirms the field is perpendicular... ...is perpendicular to the curved surface The... ...the curved surface The field through the... ...The field through the ends of the... ...the ends of the cylinder is 0... ...is 0 since the field is parallel... ...Due to a Plane of Charge E... ...be perpendicular to the plane and must... ...perpendicular to the plane and must have... ...and must have the same magnitude at... ...points equidistant from the plane Choose a... ...equidistant from the plane Choose a small... ...is perpendicular to the plane for the... ...perpendicular to the plane for the gaussian... ...the plane for the gaussian surface Field... ...Due to a Plane of Charge, cont... ...is parallel to the curved surface and... ...no contribution to the surface area from... ...curved part of the cylinder The flux... ...of the cylinder The flux through each... ...each end of the cylinder is EA... ...EA and so the total flux is... ...Due to a Plane of Charge, final... ...1040U (Physics for the biosciences) Introduction to... ...can consider of the rod and multiply... ...we consider only the x component. dy...
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