Physics Help Forum Net Electric Fields

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 Mar 21st 2018, 06:39 AM #1 Junior Member     Join Date: Nov 2017 Location: The Jurassic Forests Posts: 25 Net Electric Fields I keep getting this problem wrong- Two charges, q1 = −3.00 mC and q2 = 4.50 mC, are located at −2.00 cm and 3.00 cm on the x axis, respectively. What are the magnitude and direction of the resulting field at (a) x = −3.00 cm? (b) x = 0 cm? (c) x = 5.00 cm? My physics book doesn't bother explaining how to work field problems so I'm not sure what I'm doing wrong. I've been struggling with charge direction in electric and magnetic fields so I suppose I don't have a very good understanding of that. Anyway, in my latest attempt to solve this problem I used Coulomb's law to find individual fields and then add them. For the first one I came up with: E=k|q|/r^2 E1=k|-3e-3C|/1^2 E1=k(3e-3)/1 E1=2.697e7N/C E2=k(4.5e-3C)/6^2 E2=1.12e6 N/C 2.697e7+1.12e6= 2.8e7N/C, which the answer says is wrong. I don't understand this at all
Mar 21st 2018, 04:03 PM   #2
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 E=k|q|/r^2 E1=k|-3e-3C|/1^2 E1=k(3e-3)/1 E1=2.697e7N/C
Why 1 squared? distance should be in metres, not centimetres.

Note also the E is a vector and points to the right for q1 at a distance of -3cm from the origin. (That is it opposes the contribution from q2 at this point)

Mar 26th 2018, 04:34 PM   #3
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 Originally Posted by studiot Why 1 squared? distance should be in metres, not centimetres. Note also the E is a vector and points to the right for q1 at a distance of -3cm from the origin. (That is it opposes the contribution from q2 at this point)
Thank you!! So the charge would point to the right then?

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