Did I Approach the Electrostatic Force Derivative Correctly?

AI Thread Summary
The discussion focuses on the correct approach to deriving the electrostatic force in a physics problem. The initial solution attempts to equate the electrostatic force (Fe) with gravitational force (Fg) and breaks down the electric field into x and y components. Clarifications are sought regarding the role of tension in the system, with a distinction made between tension as a force and the sum of forces acting on an object. It is noted that tension should not simply be considered the sum of all forces, and the correct relationship involving tension and forces is emphasized. The conversation highlights the importance of precise terminology, particularly distinguishing between "derivation" and "derivative."
a_ng116
Messages
13
Reaction score
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
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
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.
 
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top