The electric field from its electric potential: semicircle

[iochoa2016]

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

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.

 

[BvU]

What makes you say that $V$ is constant ? All you bring forward is an expresssion for (I suppose?) $V$ at the point $O$ ...

$\$

 

[iochoa2016]

k = constant, lambda = constant so V = constant, isn't it?

 

[BvU]

How did you 'calculate' this $V$ ?

$\$

 

[iochoa2016]

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.

 

[haruspex]

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?

 

[iochoa2016]

found this:
along an axis pointing out the paper passing through the "O",
V = K*lambda*pi * r /(sqrt(r^2+z^2)

 

[PeroK]

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.

 

[iochoa2016]

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.

 

[bob012345]

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.

 

[bob012345]

Did the problem specifically say to use the potential or did it just ask you to derive the field?

 

[Delta2]

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).