Variation of electric field and potential along the axis of a cone

In summary, the resistance of a thin disc at a distance a from the point of the cone is R=da*J and the potential across it is V=e*H. The power dissipated in the disc is P=H*R.
  • #1
Physics lover
249
25
Homework Statement
The electric field intensity potential and rate of heat generation per unit length aare E,V and H at a section distant a from left end of a conical shaped conductor connected to a battery of emf E.Find the correct options-:
Options are in attempt at a solution
Relevant Equations
J=sigmaE
V=IR
245216


Options are at the top of page as a) b) c) d)
Answer may more than one.
Now since 'a' is distance from the smaller surface of cone so as we move along the axis area will increase,So current charge density will decrease and as we know J=sigma E,E will decrease,but V will remain constant since V=Er and r is increasing and E decreasing.Now i don't if this is correct or not and further how will i find the relation between H and a and H and E.
 
Physics news on Phys.org
  • #2
Your image misses the X axis label in d).
Please post the text of the question word for word, as another image say. Your precis is a bit vague.
If three quantities V, E and R are related by V=ER where R is increasing and E is decreasing, it does not follow that E is constant. You need to know more precisely how E and R depend on a.

Please post algebraic attempts at how E and V vary with a, perhaps as differential equations.
 
  • #3
haruspex said:
Your image misses the X axis label in d).
Please post the text of the question word for word, as another image say. Your precis is a bit vague.
If three quantities V, E and R are related by V=ER where R is increasing and E is decreasing, it does not follow that E is constant. You need to know more precisely how E and R depend on a.

Please post algebraic attempts at how E and V vary with a, perhaps as differential equations.
In option d 'a' is on x axis.So how should i proceed further.
 
  • #4
haruspex said:
Your image misses the X axis label in d).
Please post the text of the question word for word, as another image say. Your precis is a bit vague.
If three quantities V, E and R are related by V=ER where R is increasing and E is decreasing, it does not follow that E is constant. You need to know more precisely how E and R depend on a.

Please post algebraic attempts at how E and V vary with a, perhaps as differential equations.
Can you tell me which quantity will remain constant as move along axis.Is it I
 
  • #5
Physics lover said:
Can you tell me which quantity will remain constant as move along axis.Is it I
Yes. In any period of time, the flow of charge into a given disc slice of the cone must equal the flow out.
 
  • #6
haruspex said:
Yes. In any period of time, the flow of charge into a given disc slice of the cone must equal the flow out.
Ok i think i got it.since I is constant and J is increasing therefore E should increase by J=sigmaE. Am i coorect?
 
  • #7
Physics lover said:
Ok i think i got it.since I is constant and J is increasing therefore E should increase by J=sigmaE. Am i coorect?
I don’t know what J represents in this context, nor what Es are being summed.
 
  • #8
haruspex said:
I don’t know what J represents in this context, nor what Es are being summed.
J is current density.and J=sigma E is ohms law.Now can you help please.
 
  • #9
Physics lover said:
J is increasing therefore E should increase by J=sigmaE.
But here you wrote
Physics lover said:
current charge density will decrease and as we know J=sigma E,E will decrease,
Why have you switched?
 
  • #10
haruspex said:
But here you wrote

Why have you switched?
Very sorry for the first one.That was wrong J will decrease as area is increasing isn't it.
 
  • #11
Physics lover said:
Very sorry for the first one.That was wrong J will decrease as area is increasing isn't it.
Can you please help me further.I could only tell about J and I.
 
  • #12
Physics lover said:
Can you please help me further.I could only tell about J and I.
I think it will help if you are more specific about the algebraic relationship.
Consider a thin disc width da at distance a from the point of the cone. What is its resistance? What is the potential across it? How much power is dissipated in it?
 

1. What is the relationship between electric field and potential along the axis of a cone?

The electric field and potential along the axis of a cone are inversely proportional to each other. This means that as the electric field increases, the potential decreases, and vice versa.

2. How does the shape of the cone affect the variation of electric field and potential?

The shape of the cone affects the variation of electric field and potential because it determines the angle of the cone's sides, which in turn affects the distribution of charge on the surface of the cone.

3. Is there a specific formula to calculate the electric field and potential along the axis of a cone?

Yes, there is a formula that relates the electric field and potential along the axis of a cone. It is given by E = kQ/r^2 and V = kQ/r, where k is the Coulomb's constant, Q is the charge on the cone, and r is the distance from the tip of the cone to the point of interest along the axis.

4. How does the charge distribution on the cone affect the variation of electric field and potential?

The charge distribution on the cone affects the variation of electric field and potential because it determines the strength and direction of the electric field. The closer the charges are to each other, the stronger the electric field will be, resulting in a sharper variation of electric field and potential along the axis of the cone.

5. Can the electric field and potential along the axis of a cone be measured experimentally?

Yes, the electric field and potential along the axis of a cone can be measured experimentally using various methods such as equipotential mapping or using a probe to measure the electric field at different points along the axis. These measurements can then be used to verify the theoretical predictions of the variation of electric field and potential along the axis of a cone.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
318
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
2
Replies
44
Views
920
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
296
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
158
  • Introductory Physics Homework Help
Replies
1
Views
790
  • Introductory Physics Homework Help
Replies
26
Views
590
Back
Top