Homework Help: why doesn't anybody ever help me.

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

1. three point charges, q1, q2, q3, lie along the x-axis at x = 0,3,5, respectively. Calculate the magnitude and direction of the net electric force on q1 if q1 = +6 x 10^-6 C, q2 = +1.5 X 10^-6 C and q3 = -2.0 x 10^-6 C.

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 9:13pm
It'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 10:20pm
1. 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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