Calculating electrostatic force on particles

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The discussion revolves around calculating the electrostatic force on particles using the formula F=(kq1q2)/d^2. The initial attempt at solving the problem involved calculating the x-component of the force, but the result was incorrect due to a mistake in unit conversion from centimeters to meters. A participant pointed out the need to consider the signs of the forces, emphasizing that both attractive and repulsive forces must be accounted for in the calculations. The importance of accurately determining the direction of forces and the necessity of calculations over assumptions was highlighted. Overall, the conversation underscores the complexities involved in electrostatic force calculations and the need for careful attention to detail.
Jrlinton
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Homework Statement



Phy 204.PNG

Homework Equations


F=(kq1q2)/d^2

The Attempt at a Solution


So it should be fairly easy to use "parts" to come up with each answer
Part a
Because particle 1 has no pull in the x direction I should be able to ignore it when finding the x component of the force
Fx=k*q3*q4/a^2+k*q3*q2/(a*sqrt2)^2*cos45
=7.46E-4 N
This was incorrect and I did not attempt part 2 as I figured my mistake would carry over to that similar calculation
 
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Check your calculation. I get something else. Spell it out if you want it checked.
 
8.99E9*(9.5E-8)*(-9.5E-8)/(2.03E-3)+8.99E9*(9.5E-8)(-5E.7)/(4.05E-3)cos45deg
=-0.114623N
it seems that I made a mistake on my cm to m conversion and if I remember correctly this should be absolute value of the charges so
=0.114623N
 
Numerical value same as what I found.

Jrlinton said:
if I remember correctly this should be absolute value of the charges
No. You are asked for an x-component, which can be negative or positive. Make a sketch to find the right sign. What does the F in your relevant equation represent ?
 
I realized this after I sent it. Both q2 and q4 are opposite charges of q3 so they are both attractive forces. In regards to the y component q1 will repel q3 while q2 will again be attractive and by glance you can tell the effect of q1 will be greater than q2 you can assume that the answer will be negative in the y direction
 
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Assuming is one thing, calculating is better :smile:.
 
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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