Unformatted text preview: : E x = Ex / r = ( ! 180)(4)/5 = ! 144 E y = Ey / r = ( ! 180)( ! 3)/5 = + 108 Finally, write out the result with xhat and yhat and the units for electric field intensity: ! E = ( ! 144ˆ x + 108ˆ y ) N / C Notice that this equation is ready to yield the correct answer to ! F = q ! E in that any positive charge placed at the location of interest will feel a force towards the negative source charge (negative in the xdirection and positive in the ydirection), and any negative charge placed at the location of interest will be pushed away. Q = 500 nC x = 4 m y = 3 m location of interest...
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This note was uploaded on 04/05/2012 for the course PHYS 131 taught by Professor Tibbets during the Spring '11 term at Cuyamaca College.
 Spring '11
 Tibbets
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