The point charges in the figure below are located at the corners of an equilater

Question

The point charges in the figure below are located at the corners of an equilateral triangle 20.0 cm on a side. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

(a)

Find the electric field (in N/C) at the location of qa in the figure, given that qb = +12.00 µC and qc = 6.00 µC.

magnitude _________ N/C

direction __________ ° counterclockwise from the +x-axis

(b)

What is the force (in N) on qa, given that qa = +3.00 nC?

magnitude ________ N

direction ________ ° counterclockwise from the +x-axis

Explanation / Answer

Since the question is about the field, we place a point charge (1C) at the point 1 in the figure.

The filed due to qb at the position 1 in the figure is

k*qb*1/ a ² where a = 0.20 m The direction is along 3 to 1.

Its x component is k*qb* cos 60 / a ² from left to right

Its y component is k*qb* sin 60 / a ² upward
------------------------------
The filed due to qc at the position 1 in the figure is

k*qc*1/ a ² where a = 0.20 m The direction is along 1 to 2 (since the force is an attracting one.

Its x component is k*qc* cos 60 / a ² = -675*10^3 N/C from left to right (same as the previous x component)

Its y component is k*qc* sin 60 / a ² = -1169.1*10^3 N/C from down ward (opposite to the previous y component)

resultant = 1350*10^3 N/C counterclockwise from the +x-axis
--------------------------------------

The net field in the x direction is [k/ a²]* cos 60 (qb + qc) {only magnitudes of charges has to be taken here)
The net field in the y direction is [k/ a²]* sin 60 (qb - qc) {only magnitudes of charges has to be taken here) Since the magnitude of qb >qc the field is downward at the point 1.


The resultant field of these two is

[k/ a²]* {(qb + qc) ² (cos 60)² + (qb - qc)²(sin 60)²}


[9e9/0.20²]*1.e-6* {18²*0.20 +6²*0.80} = 2.18e6 N/C

Consider the coordinate axes at the point 1.

The net field is in the fourth quadrant .( y is negative and x is positive)

tan = (18 cos 60) / (6 sin 60)

= 60

The angle from the x axis measured counter- clockwise is 300

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