The electric field from its electric potential: semicircle

In summary, the electric field is not always positive at the center of a semicircle ring, which is different from what theory predicts.
  • #1
iochoa2016
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0
Homework Statement
Strugling to get the E from its V at the centre of the semicircle.
Relevant Equations
V = k*lambda* pi
pi = 3.1415
lambda = charge density
k = 1 / (4*pi*eo)
According to theory I should be able to get the Electric Field (E) from its pOtential (V) by doing the grad (V) so

E = -grad(V), however, V is contant V = k*lambda* pi which results having E =0, but this is not right. What I am missing??
see figure below
e.png
The answer should be Ex = 2*k*lambda / r

This is how I calculate V:

dV = k * dQ/r = k*r*d(theta)*lambda / r
dV = k*lambda * d(theta)
Integrating from 0 to pi

V = k* lambda * pi.
 
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  • #2
What makes you say that ##V## is constant ? All you bring forward is an expresssion for (I suppose?) ##V## at the point ##O## ...

##\ ##
 
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  • #3
k = constant, lambda = constant so V = constant, isn't it?
 
  • #4
How did you 'calculate' this ##V## ?

##\ ##
 
  • #5
This is how I calculate V:

dV = k * dQ/r = k*r*d(theta)*lambda / r
dV = k*lambda * d(theta)
Integrating from 0 to pi

V = k* lambda * pi.
 
  • #6
iochoa2016 said:
k = constant, lambda = constant so V = constant, isn't it?
You are using the expression for the potential at a specific point, but grad is the derivative wrt position. What will be the potential at a nearby point?
 
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  • #7
found this:
along an axis pointing out the paper passing through the "O",
V = K*lambda*pi * r /(sqrt(r^2+z^2)
 
  • #8
iochoa2016 said:
found this:
along an axis pointing out the paper passing through the "O",
V = K*lambda*pi * r /(sqrt(r^2+z^2)
Okay, but you can find ##\vec E## at a point directly. To obtain ##\vec E## from ##-\nabla V##, you need to find ##V## in a neighbourhood of the point. This is very unlikely to be easier.
 
  • #9
Yes, I think you are right. Just triying to use energy to work out the Field.

It was easy for the example below, but not so easy for the semicircle ring.

Screenshot 2022-02-01 225938.png
 
  • #10
iochoa2016 said:
Homework Statement:: Strugling to get the E from its V at the centre of the semicircle.
Relevant Equations:: V = k*lambda* pi
pi = 4.1516
lambda = charge density
k = 1 / (4*pi*eo)
Check your value of ##π## first unless of course you are from Indiana.:smile:
 
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  • #11
Did the problem specifically say to use the potential or did it just ask you to derive the field?
 
  • #12
The gradient is essentially the partial derivatives, and to be able to calculate a partial derivative correctly you need to know the formula of the function in an interval ##(a,b)## and not just at a specific point (0 in this case) . You need to know the formula for the potential in a subarea of the circle in this case since we have a 2D case and not just at the central point of the circle (0,0).
 
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1. What is an electric field?

An electric field is a physical field that is created by electrically charged particles. It is a force field that surrounds charged objects and exerts a force on other charged objects within its range.

2. How is the electric field related to electric potential?

The electric field is related to electric potential through the equation E = -∇V, where E is the electric field, V is the electric potential, and ∇ represents the gradient operator. This means that the electric field is the negative gradient of the electric potential.

3. What is a semicircle in relation to the electric field?

A semicircle is a half-circle shape that is often used to represent the electric field in a specific scenario. In this case, it is used to represent the electric field created by a charged object located at the center of the semicircle.

4. How is the electric field calculated from its electric potential for a semicircle?

The electric field can be calculated from its electric potential for a semicircle using the equation E = -∇V = -dV/dx, where V is the electric potential and x is the distance from the center of the semicircle. This means that the electric field at any point on the semicircle is equal to the negative derivative of the electric potential at that point.

5. What factors affect the electric field from its electric potential for a semicircle?

The electric field from its electric potential for a semicircle is affected by the magnitude of the charge at the center of the semicircle, the distance from the center of the semicircle, and the shape of the semicircle itself. Additionally, the presence of other charged objects in the surrounding area can also affect the electric field.

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