Unformatted text preview: 2 ⎢
=
⎥
2
3/ 2
2
⎢ 4πε 0 ( d / 2 ) + r ⎥ ( d / 2 )2 + r 2 4πε 0 ⎡( d / 2 )2 + r 2 ⎤
⎣
⎦
⎣
⎦ >
For r > d , we write [(d/2)2 + r2]3/2 ≈ r3 so the expression above reduces to 1 qd
 Enet  ≈
. 4πε 0 r 3 (b) From the figure, it is clear that the net electric field at point P points in the − j
direction, or −90° from the +x axis. PHY2049 Spring 2012 HW1 22 27. From symmetry, we see that the net field at P is twice the field caused by the upper semicircular charge + q = λ (π R) (and that it points downward). Adapting the steps leading to Eq. 22 21, we find 90° ⎛q⎞
λ
ˆ . Enet = 2 − ˆ
j
sin θ
= −⎜
j
4πε 0 R
ε 0π 2 R2 ⎟
⎝
⎠
−90° () (a) With R = 8.50 × 10− 2 m and q = 1.50 × 10−11 C,  Enet  = 23.8 N/C. (b) The net electric field Enet points in the − ˆ direction, or −90° counterclockwise j
from the +x axis....
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This note was uploaded on 04/02/2013 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force, Work

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