Did I Approach the Electrostatic Force Derivative Correctly?

In summary: I have attempted this problem but I am not sure if I approached it the right way. Here is a link for the diagram and the question.If anyone could check it over and point out any mistakes and tell me if I make any sense what so ever,then you would make my day.
  • #1
a_ng116
13
0
Well,I've tried attempting this problem but I am not sure if I approached it the right way. Here is a link for the diagram and the question.

[PLAIN]http://img.photobucket.com/albums/v367/crazy_cat_lady/physics/diagram1.bmp[/URL]

If anyone could check it over and point out any mistakes and tell me if I make any sense what so ever,then you would make my day.

My solution:

Fe=Fg therefore
qE=mg ------> E=mg/q

x-component: Ex¹= Ecosø
= Fe/q cosø
= mg/q cosø

y-component: Ey¹= Esinø
= Fe/q sinø
= mg/q sinø

E¹= square root of (mg/cosø)² + (mg/sinø)²

Another solution that I thought up of:

Isn't tension just the sum of all forces acting on a object...therefore:

Ft= Fe+Fg
Fe= Fg-Ft
= mgsinø- Ft
 
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  • #2
a_ng116 said:
Fe=Fg

Just wondering why this very first step? I didn't see that in the problem, although I could be blind :smile:

Also, just as an FYI, it should be a 'derivation' in the thread title, since that is the term for arriving at an equation by deriving it. A "derivative" is something else in calculus, and so that confused me initially.

If you explain that first step, I should be able to help further...

As for this question:

a_ng116 said:
Isn't tension just the sum of all forces acting on a object...

I'm afraid that the answer, in general, is no. Tension is just the name given to a force tending to 'stretch' an object, i.e. pull it apart. When referring to some object suspended by a rope, the object pulls on the rope, which pulls back up on the object, keeping it suspended. The rope is taut, not slack, so it is in tension.
 
  • #3
Yeah, the first line of your working should be Fe = Ftsin θ.

Also, how are you getting x and y components of the electric field? E is in the direction of Fe and from the diagram, Fe is only in the x direction.
 
  • #4
Hmmm...alright.It actually somewhat makes sense. There is only an x-component to the electrostatic force so starting with what nylex suggested Fe=Ftsinø,would it be true then to say that:

Fe=Ft-Fgsinø
= ma- mgsinø

My rationale for this is that Earth exerts a downward force F=-mg on the test body correct?So the test body exerts and upward force F=+mg on the earth. The upward force is caused by the tension on the thread holding the test body and this is working against gravity keeping it suspended, hence Ft=ma. Anyways...more feedback would be much appreciated even despite not making much sense before in my previous derivation. And you know, if I'm completely wrong again here.
 

FAQ: Did I Approach the Electrostatic Force Derivative Correctly?

What is the electrostatic force derivative?

The electrostatic force derivative is a mathematical concept that describes the change in electrostatic force between two charged particles as their distance or charge changes. It is used to calculate the strength of the electric field between the particles.

How is the electrostatic force derivative calculated?

The electrostatic force derivative is calculated using the Coulomb's Law formula, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The derivative is then obtained by taking the derivative of this formula with respect to the distance between the particles.

What are the units of the electrostatic force derivative?

The units of the electrostatic force derivative depend on the units used for charge and distance in the Coulomb's Law formula. If charges are measured in coulombs and distance in meters, then the units for the derivative would be N·m²/C².

What are some real-life examples of the electrostatic force derivative?

The electrostatic force derivative can be observed in many everyday situations, such as the attraction and repulsion between charged balloons, the buildup of static electricity on clothing, and the functioning of electronic devices. It is also used in industries such as electronics, semiconductors, and material science.

Why is the electrostatic force derivative important?

The electrostatic force derivative is important because it helps us understand and predict the behavior of charged particles in various situations. It also plays a crucial role in the design and functioning of electronic devices, and in the study of fundamental forces and interactions in physics.

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