Finding the angle for an electrostatic force

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The discussion centers on calculating the angle θ between the electrostatic force on a charge at the origin and the positive x-axis. The user correctly calculated the force's magnitude but struggled with determining the angle due to confusion over the signs of the components. It's emphasized that the direction of the forces, based on the charge signs, is crucial for accurate angle calculation. The user was advised to use the y-component over the x-component in the arctan function to find the correct angle. Understanding the charge interactions and their effects on force direction is essential for solving this problem accurately.
Kaani
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Homework Statement



What is the angle θ between the electrostatic force on the charge at the origin and the positive x-axis? Answer in degrees as an angle between -180 degrees and 180 degrees measured from the positive x-axis, with counterclockwise positive.

Coulomb constant is 8.98755e9

(Figure attached)

Homework Equations



F = kqQ/R2 (for answers found in previous question, which was marked correct)

The Attempt at a Solution



In the previous question I was asked to find the magnitiude of the electrostatic force on the charge at the origin (see figure). I found this to be 2.112678794e-9 by calculating the forces between the charges and the origin separately, and then using a^2+b^2=c^2.

To find the angle between the force and the x-axis I did arctan = (-9.1709694e-10)/(1.90324588e-9) and got -25.7275298 degrees. I plugged it in as both positive and negative, and both answers were wrong. I'm not sure how the "-180 to 180" part of the question changes it, and help would be appreciated. The two values I used in the tangent equation I used to find the correct answer in the question before this one, so I don't see how the angle would be wrong.

Thanks!
 

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Kaani said:

Homework Statement



What is the angle θ between the electrostatic force on the charge at the origin and the positive x-axis? Answer in degrees as an angle between -180 degrees and 180 degrees measured from the positive x-axis, with counterclockwise positive.

Coulomb constant is 8.98755e9

(Figure attached)

Homework Equations



F = kqQ/R2 (for answers found in previous question, which was marked correct)

The Attempt at a Solution



In the previous question I was asked to find the magnitiude of the electrostatic force on the charge at the origin (see figure). I found this to be 2.112678794e-9 by calculating the forces between the charges and the origin separately, and then using a^2+b^2=c^2.

To find the angle between the force and the x-axis I did arctan = (-9.1709694e-10)/(1.90324588e-9) and got -25.7275298 degrees. I plugged it in as both positive and negative, and both answers were wrong. I'm not sure how the "-180 to 180" part of the question changes it, and help would be appreciated. The two values I used in the tangent equation I used to find the correct answer in the question before this one, so I don't see how the angle would be wrong.

Thanks!

The magnitude of the force that you calculated is fine, but I think your direction is off. Take a look at the signs of the charges in the figure and pencil in the directions of the resulting forces (as small vectors) on the diagram. The signs of the forces (their directions) are important when determining the direction (angle) of the resultant. Also, make sure that you use y-component/x-component in the arctan function, otherwise you're not computing the angle that you think you are :smile:
 
gneill said:
The signs of the forces (their directions) are important when determining the direction (angle) of the resultant.

Oops... that was a stupid mistake, then. Switched the vectors around.

(picture I drew attached)

But, would the 1.903...e-9 be negitive or positive? In the equation it comes out negative, meaning it's a force of attraction. But since it's on the positive x-axis, it would have to be positive in the arctan equation, right?

So, arctan = (1.90324588e-9)/(-9.1709694e-10)
= -64.27247017
 

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Do charges of the same sign attract or repel? You should always look at the diagram and decide the direction of the resulting force. This will serve as a check on the sign of your calculated values (The signs of things in your calculation can get a bit complicated when you take the geometry of location into account. Looking at the diagram and thus knowing the required direction makes life simple :wink: ).

In this particular case you have a negative charge at the origin and a negative charge below it on the y-axis. What direction do you think the resulting force on the origin charge should be from the charge below it?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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