Voltage difference, line integral

Click For Summary
The discussion revolves around the confusion regarding the line integral of the electric field, specifically the formula for electric potential difference. The integral, represented as the dot product of the electric field E and the differential length element dL, raises questions about the direction of the dr vector and its relation to the line integral. It is clarified that dr can be defined in any direction, but the path taken may affect the potential difference in a non-conservative field. The participants also address a potential sign error in the derivation of the formula when considering the direction of dL relative to the electric field. Overall, the conversation emphasizes the importance of understanding the definitions and conventions in calculating electric potential.
kougou
Messages
80
Reaction score
0

Homework Statement

I don't understand the follow formula of the integral :

Integral of ( E dot dL) from B to A
What direction is the dr vector? Is it the direction of the line integral?

Say I want to derive the formula for electric potential due to a point in Space. E has a direction vector of ar
assume dL is pointing from infinity to A (-ar), which is opposite of the Eletric Field. And the formula ends up with a negative sign, which is not consistent with the formal formula.
Why?
 
Last edited:
Physics news on Phys.org
The integral is you described, "Integral of ( E dot dL) from B to A", is $$\int_B^A \vec{E}\cdot\vec{dL}$$... that what you meant?

What direction is the dr vector? Is it the direction of the line integral?
dr points wherever you define it to point.

The integral is along a line, which, in general, will curve in most coordinate systems.
But does the potential difference in an electric field depend on the path taken? What does the rest of the theory say?

assume dL is pointing from infinity to A (-ar), which is opposite of the Electric Field. And the formula ends up with a negative sign, which is not consistent with the formal formula.
I think you may have misplaced a minus sign.
 
Last edited:
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?

Similar threads

Replies
3
Views
1K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
64
Views
5K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K