LEARN MORE Remarks The methods used here are just like those used with Newton\'s

Question

LEARN MORE
Remarks The methods used here are just like those used with Newton's law of gravity in two dimensions.

Question Without actually calculating the electric force on q2, determine the quadrant into which the electric force vector on q2 points.
quadrant 1 where 0 < 90° quadrant 2 where 90° < 180° quadrant 3 where 180° < 270° quadrant 4 where 270° < 360° Your answer is correct.

PRACTICE IT
Use the worked example above to help you solve this problem. Consider three point charges at the corners of a triangle, as shown in the figure, where q1 = 5.50 10-9 C, q2 = -1.52 10-9 C, and q3 = 5.46 10-8 C.

EXERCISE
Using the same triangle, find the vector components of the electric force on q1 and the vector's magnitude and direction. (Use the charges given in the Practice It section.)
Fx = .....????
Fy = ?????
Go back to your diagram of the system. Use the directions of the force arrows help you to estimate the direction and relative size of the total force. Then check the vector addition. N
magnitude ???? N
direction ?????° counter-clockwise from the +x-axis

Explanation / Answer

To find the forces on q1, we need to find how each charge effects it individually using coulombs law Fe=kq1q2/r^2

so first, lets find the force of q2 on q1

Fe(q2q1) = (9e9)(5.5e-9 C)(1.52e-9 C)/32 = 8.36e-9 N.

The direction of this force is UP, since opposites attract.

Now, lets use coulombs law again to find the force that q3 does on q1

Fe(q3q1) = (9e9)(5.5e-9 C)(5.46e-8 C)/52 = 1.08e-7 N

The direction of this is repulsion, so q1 is PUSHED AWAY at an angle of 37 degrees below the negative x axis. ( they gave you the 37 degrees, you dont have to solve for it, but you could easily do it...)

Now, to find the total force, we need to take this last answer and break it down into x and y components by using sin and cos.

x force: 1.08e-7cos(37) = 8.63e-8 N LEFT

y force:  1.08e-7sin(37) = 6.5e-8 N DOWN

Total of all x forces is simply 8.63e-8 N LEFT

Total of all y forces is 8.36e-9 N UP -  6.5e-8 N DOWN = 5.66e-8 N DOWN

Now, to put it all together, use pythagorean theorem:

(8.63e-8)2 + (5.66e-8)2 = c2 therefore c = 1.032e-7 N

Finally direction. Use tan()=opposite/adjacent, so:

tan() = 5.66e-8 / 8.63e-8, and therefore = 33.3 degrees, but this is below negative x axis. To give answer with respect to positive x axis, add 180, so a total of 213.3 degrees.

To recap, the resultant force is 1.032e-7 N @ 213.3 degrees.

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