#$&*
Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Quadrant of arctan 6 over 7
** **
Set 1 Problem 7 states:
A charge of 13 `microC is located at (-16.001 m, 3 m).
If the electric field due to this charge is observed at the point (-9.001 m, 9 m), what will be its magnitude and direction?
Solution
The electric field at a point is the force per unit test charge, with the test charge located at the point. We assume a unit test charge.
We calculate the components of the electric field vector as usual:
Since the charge is positive, the force on the 1 C test charge is one of repulsion, so will be in the direction from (-16.001, 3) to (-9.001, 9).
The distance between charges is `sqrt[(-9.001--16.001)^2+( 9- 3)^2] m = 9.219 m.
The magnitude of the force is (9 x 10^9 N m^2/C^2)( 13 x 10^-6 C)(1 C)/( 9.219 m)^2 = 1376 N.
The x and y displacements from charge 13 `microC to the 1 C test charge are, respectively, 6.999999 m and 6 m, so the x and y components are in proportion 6.999999/ 9.219, and 6/ 9.219 to the force.
The x and y forces are therefore ( 6.999999/ 9.219)( 1376 N) = 1044 N and ( 6/ 9.219)( 1376 N) = 895.5 N.
Since these are the forces on a 1 C test charge, the electric field has components 1044 N/C and 895.5 N/C in the x and y directions, respectively.
We then use the standard procedure to calculate the magnitude and direction of the field vector:
The resultant force therefore has magnitude `sqrt[( 1044 N/C)^2+( 895.5 N/C)^2] = 1375 N/C.
Its direction is arctan( 6/ 6.999999)= 220.6 degrees
** **
I get the angle to be 40.6 degrees and not 220.6 degrees and I think this is so because the x and y components are positive so the angle would be in the first quadrant.
???Am I looking at this correctly???
** **
Thanks!
@&
You are correct.
arcTan(6/7) is 40.6 degrees, not 220.6 degrees. The denominator is not negative. The program generated an errouneous value.
*@