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April 1, 2015

April 1, 2015

Posted by **todi** on Thursday, February 14, 2008 at 9:03pm.

my answre is 46.8 N to the right. is that correct?

2. a charge of +2.0 x 10^-3 C is placed at the origin, and antoher charge of +4.0 x 10^-3 C is placed at x = 2m.

if a test charge, q, of +3 x 10^-9 C is placed at 1m, then determine the net electric field at that point.

my answer is 5.4 x 10^7 N/C

is that correct

- why doesn't anybody ever help me. -
**Ms. Sue**, Thursday, February 14, 2008 at 9:13pmIt's possible that an expert for these questions hasn't been online for Jiskha. Remember we are all volunteers who are leading lives outside of homework help.

- why doesn't anybody ever help me. -
**drwls**, Thursday, February 14, 2008 at 10:20pm1. In problem #1, you do not say if the locations 0, 3 and 5 are in meters or some other units. That has to be specified for a numerical answer to be obtained.

At location x=0 there is a push to the left dye to the q2-q1 repulsion and a pull to the right due to the q1-q3 attraction.

I the R distances are in meters, t he two forces are k q1q2/3^2 and k q1 q3/5^2], where k = 8.99*10^9 N/m^2C^2

This leads to a much smaller answer than yours.

2. The field at the test charge position has nothing to do with the amount of the test charge.

The E-field is k(2*10^-3)/1^2 - k(4*10^-3)/1^2

The sign changes is because charges are pushing from opposite sides. It appears you added the fields from the two charges instead iof taking the difference.

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