ch24-p038

# ch24-p038 - 38(a From the result of Problem 24-30 the...

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(b) We differentiate the potential with respect to x to find the x component of the electric field: 2 00 0 92 2 1 5 42 1 ln 44 4 ( ) (8.99 10 N m C )(43.6 10 C) (3.92 10 N m C) , ( 0.135 m) ( 0.135 m) x V Q xL Q x Q E x Lx x Lx Lx x xx L xx εε ε −− ∂∂ ⎛⎞ =− ⎜⎟ ⎝⎠ ×⋅ × × ++ ππ π or (3.92 10 N m C) || . ( 0.135 m) x E = + (c) Since 0 x E < , its direction relative to the positive x axis is 180 . ° (d) At x = d = 6.20 cm, we obtain (3.92 10 N m C) | | 0.0321 N/C. (0.0620 m)(0.0620 m 0.135 m) x E == + (e) Consider two points an equal infinitesimal distance on either side of P 1 , along a line that is perpendicular to the x axis. 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.

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