Electricity, Direction of Charge Problem

[Bound]

Some background: This was a quiz I recently had in which I lost a few points for missing an "x & y component" for one of the directions.

My thinking is that because the problem was not actually set up with a specified order of charges there are a few different answers you could get.

I intend to bring this up with the teacher, but thought I'd see what you guys have to say as well.

Homework Statement

Three charges are arranged in the shape of an equilateral triangle. (Note- no diagram is given)

The charge magnitudes are as follows:
q₁=+4*10^-6C, q₂=-6.5*10^-6C, q₃=+2.1*10^-6C

The distance between each charge is 2m

Calculate the electric force on q₃.

Homework Equations

F = (kq₁q₂) / d²

The Attempt at a Solution

I set my triangle up as: (see attached picture)

Calculated force of 1 on 3:
F = .0189 N

X-component of F₁₃:
F_x=(.0189)(sin30)
=.00945 N

Y-component of F₁₃:
F_y=(.0189)(cos30)
=.0163676801 N

*Marked Correct*

Calculated force of 2 on 3:
F = -.0307125 N

*Marked incorrect for not separating into x & y components*

Would this not be correct though? The way I set my triangle up has 2 on a straight horizontal line across from 3!

My further work is just adding up all of the components of everything for a final answer of:
Net Force on q3 = .0268 N @ 37.6deg SofW

Thanks for any help!

 

[gneill]

Some background: This was a quiz I recently had in which I lost a few points for missing an "x & y component" for one of the directions.

My thinking is that because the problem was not actually set up with a specified order of charges there are a few different answers you could get.

I intend to bring this up with the teacher, but thought I'd see what you guys have to say as well.

Homework Statement

Three charges are arranged in the shape of an equilateral triangle. (Note- no diagram is given)

The charge magnitudes are as follows:
q₁=+4*10^-6C, q₂=-6.5*10^-6C, q₃=+2.1*10^-6C

The distance between each charge is 2m

Calculate the electric force on q₃.

Homework Equations

F = (kq₁q₂) / d²

The Attempt at a Solution

I set my triangle up as: (see attached picture)

Calculated force of 1 on 3:
F = .0189 N

X-component of F₁₃:
F_x=(.0189)(sin30)
=.00945 N

Y-component of F₁₃:
F_y=(.0189)(cos30)
=.0163676801 N

*Marked Correct*

Calculated force of 2 on 3:
F = -.0307125 N

*Marked incorrect for not separating into x & y components*

Would this not be correct though? The way I set my triangle up has 2 on a straight horizontal line across from 3!

My further work is just adding up all of the components of everything for a final answer of:
Net Force on q3 = .0268 N @ 37.6deg SofW

Thanks for any help!

You must always specify units and direction for any vector quantities, otherwise points will be deducted. It's not enough to assume that unspecified directions imply a direction along the positive x-axis (or Easterly), unless you so specify in your work.

 

[Bound]

So would I just have had to specify xx.xx N [0deg. E] to be correct?

Or are you implying that there is some arbitrary angle in which that force can be broken into a y-component, if so I don't have the first idea on how to find that angle.

 

[gneill]

So would I just have had to specify xx.xx N [0deg. E] to be correct?

That would work.

Or are you implying that there is some arbitrary angle in which that force can be broken into a y-component, if so I don't have the first idea on how to find that angle.

All planar vectors have two components, even if one of them happens to have a zero magnitude. When one of the components happens to be zero you can drop it so long as you've specified the direction for the remaining component. It would not be incorrect to include the zero magnitude component as, for example, 0.00 N [North].

In your work for the other forces you explicitly broke them out as F_x and F_y. You could have done the same for F₂₃ and written

For F₂₃:
F_x = xx.xx N
F_y = 0.00 N