(b) We differentiate the potential with respect to xto find the xcomponent of the electric field: 200092215421ln444()(8.99 10 N m C )(43.6 10C)(3.92 10N m C),(0.135 m)(0.135 m)xVQxLQxQExLxxLx Lx xxx Lxxεεε−−∂∂−−⎛⎞⎛⎞=−−⎜⎟⎜⎟−−⎝⎠⎝⎠×⋅××⋅++πππor (3.92 10N m C)||.(0.135 m)xE−=+(c) Since0xE<, its direction relative to the positive xaxis is 180 .°(d) At x = d= 6.20 cm, we obtain (3.92 10N m C)||0.0321 N/C.(0.0620 m)(0.0620 m 0.135 m)xE−==+(e) Consider two points an equal infinitesimal distance on either side of P1, along a line that is perpendicular to the xaxis. The difference in the electric potential divided by their separation gives the transverse component of the electric field. Since the two points are
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This note was uploaded on 10/28/2010 for the course PHY 364 taught by Professor Ve during the Fall '10 term at Valley Forge Christian.