Resultant Force on -2x10^-6 C Point Charge Near 3 & -4x10^-6 C

[wheybags]

Homework Statement

Determine the resultant force acting on a point charge of -2 x 10 ^-6 C situated at the origin of a rectangular coordinate system in the vicinity of point charges 3x10^-6 C and -4x10^-6 C at distances 0.12 m along the positive x-axis and 0.08m along the half-line x=y where x,y>0 respectively.

Homework Equations

Coulomb's law

The Attempt at a Solution

I used Coulomb's law twice, taking first the first and second points, the the first and third points, and got -3.75 and 11.25 respectively.
However, the answer is supposedly 14.15 < 34.20°, which I am nowhere near to, and do not even understand.
Help greatly appreciated.

 

[Andrew Mason]

Homework Statement

Determine the resultant force acting on a point charge of -2 x 10 ^-6 C situated at the origin of a rectangular coordinate system in the vicinity of point charges 3x10^-6 C and -4x10^-6 C at distances 0.12 m along the positive x-axis and 0.08m along the half-line x=y where x,y>0 respectively.

Homework Equations

Coulomb's law

The Attempt at a Solution

I used Coulomb's law twice, taking first the first and second points, the the first and third points, and got -3.75 and 11.25 respectively.

This is the correct magnitude of the two force vectors (what are the units?). What do you get for the magnitude and direction of the vector sum of these two force vectors?

However, the answer is supposedly 14.15 < 34.20°, which I am nowhere near to, and do not even understand.
.

That answer does not appear to be correct. You can easily tell the general direction of the resultant force - it has to be below the line x=y where x,y<0

AM

 

[wheybags]

What do you get for the magnitude and direction of the vector sum of these two force vectors?

I don't know how to do that... Sorry, I'm pretty awful at this.
Could anyone perhaps do out a step-by-step solution? I could use it as an example for other questions.

 

[Piyu]

One thing i like to do when it comes down to charge and vectors is to first find the force Qqk/r then later multiply it by its respective i j k multipliers

Eg. F(cos a i + sin a J).

Then i do this for both charges and sum up the respective vectors.

 

[Andrew Mason]

I don't know how to do that... Sorry, I'm pretty awful at this.
Could anyone perhaps do out a step-by-step solution? I could use it as an example for other questions.

Ok. The force from the charge that is .12 m away on the x-axis is +3.75 N attractive so it is in the direction of the +x axis. Draw an arrow from the origin with a length of 3.75 N pointing in the positive x direction. We write that vector as: (3.75,0)

The force that is .08 m away on the line x = y, which is 45 degrees above the x axis, is -11.25 N, repulsive, so it is away from that charge ie. in a direction of 180+45 = 225 degrees. The x component of that force is 11.25cos 225 = -7.95 N. The y component is the same: 11.25 sin 225 = -7.95 N. We write that vector as: (-7.95, -7.95) .

To find the resultant vector, add them: (3.75,0) + (-7.95, -7.95). To do this, add the x components to find the x component of the sum of those two vectors: x component = 3.75-7.95 = -4.20 N. The y component of the resultant is -7.95 + 0 = -7.95 N.

So the result is a vector (-4.2,-7.95).

Can you work out the length and the direction of that vector?

AM

 

[wheybags]

I don't understand why you reversed the sign of the two charges.

Also:

The force that is .08 m away on the line x = y, which is 45 degrees above the x axis, is -11.25 N, repulsive, so it is away from that charge ie. in a direction of 180+45 = 225 degrees. The x component of that force is 11.25cos 225 = -7.95 N. The y component is the same: 11.25 sin 225 = -7.95 N. We write that vector as: (-7.95, -7.95) .

Why would the x and y component be 11.25 cos 225?
Could one not do it like this?

[PLAIN]http://img146.imageshack.us/img146/7062/grapheg.png

sin 45 = o/h = y/0.8 = 1/sqrt(2)

=> y = 0.8/sqrt(2) = 0.565

But obviously this is a different answer.

Also:

Can you work out the length and the direction of that vector?

The question was to determine the resultant force acting on the point, not the sum of the two other vectors, or is that the same thing?

 

[Sourabh N]

I don't understand why you reversed the sign of the two charges.

Since the charge at origin is negative, the force by the charge lying on x-axis will be attractive (and since it lies on positive side of x-axis, the direction of force on charge at origin is positive), and repulsive by the charge lying on x=y (by similar argument as above, force on charge at origin in the negative x,y direction).

Why would the x and y component be 11.25 cos 225?
Could one not do it like this?

[PLAIN]http://img146.imageshack.us/img146/7062/grapheg.png

sin 45 = o/h = y/0.8 = 1/sqrt(2)

=> y = 0.8/sqrt(2) = 0.565

But obviously this is a different answer.

Not sure what you are trying to say here.

The question was to determine the resultant force acting on the point, not the sum of the two other vectors, or is that the same thing?

Yes, they are indeed the same thing.

 

[wheybags]

Not sure what you are trying to say here.

I'm asking why the method I proposed is not a valid solution, and why the the one that AM posted is.

 

[Andrew Mason]

I'm asking why the method I proposed is not a valid solution, and why the the one that AM posted is.

You are taking the y component of the displacement for some reason that you do not explain.

You have to take the x and y components of the FORCE vector not the displacement vector. The length of the FORCE vector is inversely proportional to the square of the length of the displacement vector.

We represent each force by a vector. The vector's length is proportional to the force and points in the direction of the force. To add forces, we add the force vectors using the vector addition method (put two force vectors tail to head and then measure the length and direction of the vector from the tail of the first to the head of the second). That is all I have done.

AM