Respuesta :
a) 36.3 N
b) [tex]204.4^{\circ}[/tex] clockwise from positive x-direction
c) 43.7 N
d) [tex]-10.9^{\circ}[/tex]
e) 25.3 N
f) [tex]112.8^{\circ}[/tex]
Explanation:
a)
The electrostatic force between two charges is given by
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1, q2 are the two charges
r is the distance between the two charges
For charge A, we have:
[tex]q_A=+3.00\mu C = +3.00\cdot 10^{-6}C\\q_B=+6.00\mu C=+6.00\cdot 10^{-6}C\\q_C=+2.00 \mu C=+2.00\cdot 10^{-6}C[/tex]
The distances are:
[tex]r_{AB}=7.00 cm =0.07 m\\r_{AC}=6.00 cm = 0.06 m[/tex]
The force of charge B on charge A is
[tex]F_{AB}=k\frac{q_A q_B}{r_{AB}^2}=(9\cdot 10^9)\frac{(3.00\cdot 10^{-6})(6.00\cdot 10^{-6})}{0.07^2}=33.1 N[/tex]
Both charges are positive, so the force is repulsive, so towards the negative x-direction (because charge A is at the origin and charge B is on the positive x-axis)
The force of charge C on charge A is
[tex]F_{AC}=k\frac{q_A q_C}{r_{AC}^2}=(9\cdot 10^9) \frac{(3.00\cdot 10^{-6})(2.00\cdot 10^{-6})}{0.06^2}=15 N[/tex]
And since charge C is positive and located on the positive y-axis, this force is repulsive, and so towards negative y-direction.
So the two forces are perpendicular, so the net force is:
[tex]F_A=\sqrt{F_{AB}^2+F_{AC}^2}=\sqrt{33.1^2+15^2}=36.3 N[/tex]
b)
Force A has two components: one in the negative x-direction, one in the negative y-direction. Therefore, the force points towards the 3rd quadrant (south-west direction).
We can find the angle by using:
[tex]\theta = tan^{-1}(\frac{F_{AC}}{F_{AB}})=tan^{-1}(\frac{15}{33.1})=24.4^{\circ}[/tex]
This the angle with respect to the negative x-direction: therefore, if we want to express the angle as clockwise angle from positive direction, it is
[tex]\theta=180^{\circ}+24.4^{\circ}=204.4^{\circ}[/tex]
c)
The force exerted by charge A on charge B is already calculated in part a:
[tex]F_{AB}=33.1 N[/tex]
And charge B, it points towards positive x-direction.
The distance between charge B and C is:
[tex]r_{BC}=\sqrt{r_{AB}^2+r_{AC}^2}=\sqrt{0.07^2+0.06^2}=0.092 m[/tex]
So the force between charge B and C is
[tex]F_{BC}=k\frac{q_B q_C}{r_{BC}^2}=(9\cdot 10^9)\frac{(6.00\cdot 10^{-6})(2.00\cdot 10^{-6})}{0.092^2}=12.8 N[/tex]
To find the net force we have to resolve the force BC in both directions. The angle between force BC and the positive x-axis is
[tex]\theta=tan^{-1}(\frac{6.00}{7.00})=40.6^{\circ}[/tex]
In the south-east direction (so, below the positive x-axis). So the components are
[tex]F_{BCx}=F_{BC}cos\theta=(12.8)(cos (-40.6^{\circ}))=9.8 N\\F_{BCy}=F_{BC}sin \theta =(12.8)(sin(-40.6^{\circ}))=-8.3 N[/tex]
So the net force on the two axis are:
[tex]F_x=F_{AB}+F_{BCx}=33.1+9.8=42.9 N\\F_y=F_{BCy}=-8.3 N[/tex]
So the net force on charge B is
[tex]F_B=\sqrt{F_x^2+F_y^2}=\sqrt{42.9^2+(-8.3)^2}=43.7 N[/tex]
d)
The direction of the force on B is given by
[tex]\theta=tan^{-1}(\frac{F_y}{F_x})[/tex]
and substituting,
[tex]\theta=tan^{-1}(\frac{-8.3}{42.9})=-10.9^{\circ}[/tex]
e)
The forces on charge C from the other two charges have been already calculated:
[tex]F_{AC}=15 N\\F_{BC}=12.8 N[/tex]
The force AC points upward along the y-positive direction, while the force BC points in the north-west direction, so we can resolve it in the two directions:
[tex]F_{BCx}=-F_{BC}cos\theta=(12.8)(cos (40.6^{\circ}))=-9.8 N\\F_{BCy}=F_{BC}sin \theta =(12.8)(sin(40.6^{\circ}))=8.3 N[/tex]
So the components of the resultant force on C are
[tex]F_x=F_{BCx}=-9.8 N\\F_y=F_{AC}+F_{BCy}=15+8.3=23.3 N[/tex]
So the magnitude of the resultant force is
[tex]F_C=\sqrt{F_x^2+F_y^2}=\sqrt{(-9.8)^2+(23.3)^2}=25.3 N[/tex]
f)
The direction of the force on C is given by
[tex]\theta=tan^{-1}(\frac{F_y}{F_x})[/tex]
Substituting, we find
[tex]\theta =tan^{-1}(\frac{23.3}{9.8})=67.2^{\circ}[/tex]
And this angle is above the negative x-axis: so, the angle clockwise with respect to the positive x-direction is
[tex]\theta=180^{\circ}-67.2^{\circ}=112.8^{\circ}[/tex]
