Electric Fields: Supporting a 4kg Chicken

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Homework Help Overview

The discussion revolves around the application of electric fields to support a 4 kg chicken, exploring the relationship between electric force and gravitational force in a static scenario. Participants are examining the implications of charge required to balance the forces acting on the chicken in an electric field of strength approximately 3 x 10^6 N/C.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the balance of forces, specifically how the upward electric force must equal the downward gravitational force for the chicken to remain stationary. There are attempts to calculate the required charge and questions about the implications of the charge's polarity.

Discussion Status

Some participants have provided calculations and reasoning regarding the forces involved, while others have raised questions about the correctness of these calculations and the assumptions made. There is an ongoing exploration of the relationship between gravitational and electric forces, with no explicit consensus reached on the interpretations presented.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for discussion. There are also references to specific values and assumptions regarding gravitational acceleration and electric field strength that are being questioned and clarified throughout the thread.

Lalitha
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Hello Sir ,
i am new to the site and so is not familar of that but i have question and i found u online so wish u to help me with that.

Electric forces are used to accelerate small particles (like ink particles in a printers) but consider electrostatic action on macroscopic objects .Suppose that a 4 kg chicken were to be supported by a vertically upwardly directed electric field .Since the amximum E strength in air is about 3*10^6.The net charge on each chicken would be unreasonably large since it must be least.

a) 1.3*10^-6

b)1.3*10^-5

c)-1.3*10^-5

d)7.7*10^4

none
 
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Please show some work.

In a static situation, the upward electric force must be equal and opposite the weight of the mass being suspended.
 
thanks a lot sir i got that.I am very Thankful for your help .
 
Astronuc said:
Please show some work.

In a static situation, the upward electric force must be equal and opposite the weight of the mass being suspended.

as we are said that the electric force is equal to mass but in oppostie direction then the force that makes the body stantionary in an electric field is F=-mg=-4*10 (g=9.8m/sec^2=10 app),


And as E=3*10^6 and using the realtion that F=qE

we get q=F/E=-4*10/(3*10^6)=-1.3*10^-5
 
Lalitha said:
as we are said that the electric force is equal to mass but in oppostie direction then the force that makes the body stantionary in an electric field is F=-mg=-4*10 (g=9.8m/sec^2=10 app),


And as E=3*10^6 and using the realtion that F=qE

we get q=F/E=-4*10/(3*10^6)=-1.3*10^-5
Be careful here.

The force of gravity is down so the force of the electric field must be up, so Fe = -Fg, so if Fg = -40N, the Fe = 40 N. What does that indicate about the polarity of the charge?
 
EDIT: beaten by Astro :biggrin:
 
Astronuc said:
Be careful here.

The force of gravity is down so the force of the electric field must be up, so Fe = -Fg, so if Fg = -40N, the Fe = 40 N. What does that indicate about the polarity of the charge?

Sir here as we consider the force of gravity so it need to be +40N NA SIR as we have given that m=4kg,
and f=mg =4*10=40,
and Fe=-Fg=-40.

Is that not right Sir ?
 
Hello Sir Plz let me know am i correct or going wrong there mean as fg=mg=4*10=40 and Fe=-Fg=-40N
 
Fe=-Fg= 40N

The gravitational force is downward, and by convention, that is negative.

Acceleration is a vector as is force. By convention a downward vector is negative.

The acceleration of gravity is [tex]\vec{a}\,=\,-9.81 \hat{j}\,m/s^2[/tex], where [tex]\hat{j}[/tex] is the unit vector in the vertical orientation and is taken as positive.
 
  • #10
Ok thanks a lot for that sir .I got it now .
 
  • #11
Kurdt said:
EDIT: beaten by Astro :biggrin:

The plates are connected to a battery and then the connection is removed so the two plates are left charged and isolated. If a dielectric material such as a sheet of glass is inserted between the plates:

a. no work is done and the energy stored stays constant.
b. work is done, increasing the amount of stored energy in the capacitor.
c. work is done, decreasing the amount of energy stored in the capacitor.
 
  • #12
Plz can anybody help me with this one
 

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