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
2 answers
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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.
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.
0 answers