Finding Magnitude and Direction of Force on a Charge

[cheerspens]

Homework Statement

Three charges are arranged as shown in the picture I attached. Find the magnitude and direction of the electrostatic force on the charge at the origin.

Homework Equations

I know that Coulomb's Law must be applied.

The Attempt at a Solution

I've drawn a force diagram with F_CB pointing to the left and F_AB pointing down. I've also broke the vectors into components using Coulomb's Law and added them together. By doing this, I got 2.7 x 10^-5 \hat{x} + 0 \hat{y}. From here what do I do to get an answer?

I know the answer is supposed to be 1.38 x 10^-5 at 77.5 degrees so I need help with how to get there.

 

[Doc Al]

I've drawn a force diagram with F_CB pointing to the left and F_AB pointing down. I've also broke the vectors into components using Coulomb's Law and added them together. By doing this, I got 2.7 x 10^-5 \hat{x} + 0 \hat{y}.

You must have made a mistake, since the resultant must have a component in the y direction. Show how you computed F_CB and F_AB.

 

[cheerspens]

F_CB=[(9x10⁹|(6x10^-9)(5x10^-9)|) / (.3²)][cos 180]\widehat{x} + [(9x10⁹|(6x10^-9)(5x10^-9)|) / (.3²)][sin 180]\widehat{y}

F_AB=[(9x10⁹|(-3x10^-9)(5x10^-9)|) / (.1²)][cos 270]\widehat{x} +[(9x10⁹|(-3x10^-9)(5x10^-9)|) / (.1²)][sin 270]\widehat{y}

 

[Doc Al]

F_CB=[(9x10⁹|(6x10^-9)(5x10^-9)|) / (.3²)][cos 180]\widehat{x} + [(9x10⁹|(6x10^-9)(5x10^-9)|) / (.3²)][sin 180]\widehat{y}

F_AB=[(9x10⁹|(-3x10^-9)(5x10^-9)|) / (.1²)][cos 270]\widehat{x} +[(9x10⁹|(-3x10^-9)(5x10^-9)|) / (.1²)][sin 270]\widehat{y}

OK, now simplify these results by evaluating those trig functions.

 

[cheerspens]

I did and that's how I got 2.7x10^-9\hat{x} + 0\hat{y}
Was my set up right and I went wrong somewhere in the math? I was always thinking I set it up wrong or wasn't using the right values.

 

[Doc Al]

Give me your values for sin(180), cos(180), sin(270), and cos(270).

 

[cheerspens]

sin(180) = 0
cos(180) = -1
sin(270) = -1
cos(270)=0

 

[Doc Al]

sin(180) = 0
cos(180) = -1
sin(270) = -1
cos(270)=0

Good. Now simplify your equations in post #3 accordingly.

 

[Lobezno]

What's the point of incorporating Trig functions? If you compute them individually then you get two separate forces, one along the y-axis and one along the x axis. Remember also that the force between charges A and B is repulsive, so the force you drew downwards is meant to be upwards.

 

[Doc Al]

What's the point of incorporating Trig functions? If you compute them individually then you get two separate forces, one along the y-axis and one along the x axis.

While you don't need to use trig functions, there's nothing wrong with using them.

Remember also that the force between charges A and B is repulsive, so the force you drew downwards is meant to be upwards.

Charges A and B are oppositely charged, thus the force is attractive.

 

[Lobezno]

C and B, sorry. They'll produce an opposite charge.