CoulombPoints - E x = Ex r = 180(4/5 = 144 E y = Ey r = 180...

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Coulomb’s Law for Point Charges Suppose you wish to find the electric field vector at a given location due to the proximity of an electrostatic point-charge Q as shown in the diagram to the right. First, find the magnitude of the electric field using E = kQ/r 2 . It should be obvious that r can be found using the Pythagorean theorem: r = x 2 + y 2 = 16 + 9 = 5 E = kQ / r 2 = 9 ! 10 9 ( ) " 500 ! 10 " 9 ( ) / 5 2 ( ) = " (9)(20) = " 180 Next, for this problem set-up, recognize that there is no need to compute an angle and then take the sine and cosine components. Instead, the x -component of the vector is simply the magnitude times x/r and the y -component is the magnitude times y/r
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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 x-hat and y-hat 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 x-direction and positive in the y-direction), 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.

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