Question

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

Answer 1

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)`

Step-by-step explanation:

a)

The electrostatic force between two charges is given by

`F=k(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(q_A q_B)/(r_(AB)^2)=(9\cdot 10^9)((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(q_A q_C)/(r_(AC)^2)=(9\cdot 10^9) ((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}=√(33.1^2+15^2)=36.3 N`

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:

`\theta = tan^(-1)((F_(AC))/(F_(AB)))=tan^(-1)((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)`

c)

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}=√(0.07^2+0.06^2)=0.092 m`

So the force between charge B and C is

`F_(BC)=k(q_B q_C)/(r_(BC)^2)=(9\cdot 10^9)((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)((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=√(F_x^2+F_y^2)=√(42.9^2+(-8.3)^2)=43.7 N`

d)

The direction of the force on B is given by

`\theta=tan^(-1)((F_y)/(F_x))`

and substituting,

`\theta=tan^(-1)((-8.3)/(42.9))=-10.9^(\circ)`

e)

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`

So the components of the resultant force on C are

`F_x=F_(BCx)=-9.8 N\\F_y=F_(AC)+F_(BCy)=15+8.3=23.3 N`

So the magnitude of the resultant force is

`F_C=√(F_x^2+F_y^2)=√((-9.8)^2+(23.3)^2)=25.3 N`

f)

The direction of the force on C is given by

`\theta=tan^(-1)((F_y)/(F_x))`

Substituting, we find

`\theta =tan^(-1)((23.3)/(9.8))=67.2^(\circ)`

And this angle is above the negative x-axis: so, the angle clockwise with respect to the positive x-direction is

`\theta=180^(\circ)-67.2^(\circ)=112.8^(\circ)`