Is the Electric Field Equation for a Coaxial Cable Correct?

[CL39]

Homework Statement

GIVEN:
a=radius of inner electrode = 0.05 cm
b=radius of outer electrode = 0.5 cm
Vo = 130 V

QUESTION: What is the electric field at surface of the inner electrode in units of V/m?

Homework Equations

Given equations
Potential at radius r b/w inner&outer electrodes is
V=Vo*ln(r/a) (Eq 1)

Elect field at r is
E(V/m) = (1/r)*[V/ln(b/a)] Eq (2)

The Attempt at a Solution

First in finding the V at surface of inner electrode,
the value of r is same as a because it's at surface, so the ln (r/a) term
becomes zero. So that is where I am stuck. Are the equations I am given correct?

 

[collinsmark]

Homework Statement

GIVEN:
a=radius of inner electrode = 0.05 cm
b=radius of outer electrode = 0.5 cm
Vo = 130 V

QUESTION: What is the electric field at surface of the inner electrode in units of V/m?

Homework Equations

Given equations
Potential at radius r b/w inner&outer electrodes is
V=Vo*ln(r/a) (Eq 1)

Elect field at r is
E(V/m) = (1/r)*[V/ln(b/a)] Eq (2)

The Attempt at a Solution

First in finding the V at surface of inner electrode,
the value of r is same as a because it's at surface, so the ln (r/a) term
becomes zero. So that is where I am stuck. Are the equations I am given correct?

Your Eq. 1 doesn't look right to me. $V$ should be equal to $V_0$ when $r = a$. So something is not right with that equation.

If you're curious (and Ill give this because this thread is so old), I think that equation should be something like

$V = \frac{-V_0 \ln(r)}{\ln \left( \frac{b}{a} \right)} + \frac{V_0 \ln(b)}{\ln \left(\frac{b}{a}\right)} \ \ \ (b \gt r \gt a)$

But you shouldn't need Eq. 1 to find the electric field.

Eq. 2 looks good to me though. Doesn't that equation give you the answer you want?