HW1-phy2049-Spring-2012-solution

The acceleration of the proton is ap eemp and the

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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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