Physics, 02.03.2020 17:29 kenoknox

Point charge A with a charge of +3.00 μC is located at the origin. Point charge B with a charge of +6.00 μC is located on the x axis at x = 7.00 cm. And point charge C with a charge of +2.00 μC is located on the y axis at y = 6.00 cm. What is the net force (magnitude and direction) exerted on each charge by the others?

a. charge A magnitude

b. charge A direction

c. charge B magnitude

d. charge B direction

e. charge C magnitude

f. charge C direction

Answers: 3

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Answers

Answer from: MendesArmy333

a) 36.3 N

b) $204.4^{\circ}$ clockwise from positive x-direction

c) 43.7 N

d) $-10.9^{\circ}$

e) 25.3 N

f) $112.8^{\circ}$

Explanation:

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

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:

$q_A=+3.00\mu C = +3.00\cdot 10^{-6}C\\q_B=+6.00\mu C=+6.00\cdot 10^{-6}C\\q_C=+2.00 \mu C=+2.00\cdot 10^{-6}C$

The distances are:

$r_{AB}=7.00 cm =0.07 m\\r_{AC}=6.00 cm = 0.06 m$

The force of charge B on charge A is

$F_{AB}=k\frac{q_A q_B}{r_{AB}^2}=(9\cdot 10^9)\frac{(3.00\cdot 10^{-6})(6.00\cdot 10^{-6})}{0.07^2}=33.1 N$

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

$F_{AC}=k\frac{q_A q_C}{r_{AC}^2}=(9\cdot 10^9) \frac{(3.00\cdot 10^{-6})(2.00\cdot 10^{-6})}{0.06^2}=15 N$

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:

$F_A=\sqrt{F_{AB}^2+F_{AC}^2}=\sqrt{33.1^2+15^2}=36.3 N$

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:

$\theta = tan^{-1}(\frac{F_{AC}}{F_{AB}})=tan^{-1}(\frac{15}{33.1})=24.4^{\circ}$

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

$\theta=180^{\circ}+24.4^{\circ}=204.4^{\circ}$

The force exerted by charge A on charge B is already calculated in part a:

$F_{AB}=33.1 N$

And charge B, it points towards positive x-direction.

The distance between charge B and C is:

$r_{BC}=\sqrt{r_{AB}^2+r_{AC}^2}=\sqrt{0.07^2+0.06^2}=0.092 m$

So the force between charge B and C is

$F_{BC}=k\frac{q_B q_C}{r_{BC}^2}=(9\cdot 10^9)\frac{(6.00\cdot 10^{-6})(2.00\cdot 10^{-6})}{0.092^2}=12.8 N$

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

$\theta=tan^{-1}(\frac{6.00}{7.00})=40.6^{\circ}$

In the south-east direction (so, below the positive x-axis). So the components are

$F_{BCx}=F_{BC}cos\theta=(12.8)(cos (-40.6^{\circ}))=9.8 N\\F_{BCy}=F_{BC}sin \theta =(12.8)(sin(-40.6^{\circ}))=-8.3 N$

So the net force on the two axis are:

$F_x=F_{AB}+F_{BCx}=33.1+9.8=42.9 N\\F_y=F_{BCy}=-8.3 N$

So the net force on charge B is

$F_B=\sqrt{F_x^2+F_y^2}=\sqrt{42.9^2+(-8.3)^2}=43.7 N$

The direction of the force on B is given by

$\theta=tan^{-1}(\frac{F_y}{F_x})$

and substituting,

$\theta=tan^{-1}(\frac{-8.3}{42.9})=-10.9^{\circ}$

The forces on charge C from the other two charges have been already calculated:

$F_{AC}=15 N\\F_{BC}=12.8 N$

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:

$F_{BCx}=-F_{BC}cos\theta=(12.8)(cos (40.6^{\circ}))=-9.8 N\\F_{BCy}=F_{BC}sin \theta =(12.8)(sin(40.6^{\circ}))=8.3 N$