Potential due to a line charge

[Buffu]

Homework Statement

A line charge with uniform charge distribution and linear charge density $\lambda$ lie along z-axis from $(0,0,-d)$ to $(0,0,+d)$. Find electric field due to this charge along z, x-axis and potential at $(0,0,2d)$, $(\sqrt{3}d, 0,0)$ and $(3d/2, 0 , d)$.

Homework Equations

The Attempt at a Solution

$\displaystyle dE_z = \dfrac{dQ}{R^2} = {\lambda dR \over R^2}$

Hence, $\displaystyle E_z = \int^{z+d}_{z-d} {\lambda dR \over R^2} = {2\lambda d \over z^2 - d^2}$

Where $z$ is a point along z-axis.Now for potential
$\displaystyle \phi_z = -\int^z_0 E_z dz = -\int^z_0 {2\lambda d \over z^2 - d^2} dz = -\lambda\ln \left| d^2 (z-d) \over z + d\right|$

Now for x-axis,

$\displaystyle dE_x = {dQ \cos \theta \over R^2} = {\lambda dz \cos \theta \over R^2} = {\lambda dz \cos \theta \over x^2 + z^2} = {\lambda x \over (x^2 + z^2)^{3/2}}dz$

So, $\displaystyle E_x = \int^{d}_{-d}{\lambda x \over (x^2 + z^2)^{3/2}} dz = {2\lambda d\over x(d^2 + x^2)^{1/2}}$

Now for potential,

$\displaystyle \phi_x = -\int^x_0 {2\lambda d \over x(x^2 + d^2)^{1/2}} = -\lambda \ln\left| \sqrt{d^2 +x^2} - d \over \sqrt{x^2 + d^2 } + d\right|$

Now we need to find potential at $(0,0,2d)$, $(\sqrt{3}d, 0,0)$ and $(3d/2, 0 , d)$,

Plugging $(0,0,2d)$ in $\phi_z$ I got $\lambda \ln 3$.

Plugging $(\sqrt{3} \times d,0,0)$ in $\phi_x$ I got $\lambda \ln 3$.

So far so good,

For potential at $(3d/2, 0 , d)$, I plugged this coordinate in

$\displaystyle \phi (x,0,z) =\phi_x + \phi_z = -\lambda \ln\left| \sqrt{d^2 +x^2} - d \over \sqrt{x^2 + d^2 } + d\right| + -\lambda\ln \left| d^2 (z-d) \over z + d\right|$

Now as you see for $z = d$ the expression on the far right is undefined because $\ln 0$ is undefined.

What to do now ? I got every answer correct except this one. The given answer is $\lambda\ln 3$.

What did I do wrong ?

 

[haruspex]

$\displaystyle \phi (x,0,z) =\phi_x + \phi_z$

Why?
Your φ_x is shorthand for φ(x,0,0). Similarly φ_z.

 

[Buffu]

Why?
Your φ_x is shorthand for φ(x,0,0). Similarly φ_z.

I guess because $\phi$ is scalar, so it adds up like a scalar.

$\displaystyle \phi = \int \vec E \cdot d\vec s = \int E_x dx + E_ydy + E_z dz = \phi_x + \phi_z$

I took $\phi_y = 0$ because we are not considering it.

Is this correct ?

 

[haruspex]

I guess because $\phi$ is scalar, so it adds up like a scalar?

It is a potential, so adds up like a potential. The potential at B is the potential at A plus the potential difference from A to B.
Don't know what you mean about its being scalar. You seem to be assuming it is linear, i.e. φ(x+x',y,z)=φ(x,y,z)+φ(x',0,0).
(Edited... wrote φ(x',y,z) by mistake.)

 

[Buffu]

It is a potential, so adds up like a potential. The potential at B is the potential at A plus the potential difference from A to B.
Don't know what you mean about its being scalar. You seem to be assuming it is linear, i.e. φ(x+x',y,z)=φ(x,y,z)+φ(x',y,z).

So $\displaystyle \phi(x, 0, z) = \phi_x + \int_{(x, 0 , 0 )}^{(x,0,z)} \vec E \cdot d\vec s$ is correct ?

 

[haruspex]

So $\displaystyle \phi(x, 0, z) = \phi_x + \int_{(x, 0 , 0 )}^{(x,0,z)} \vec E d\vec s$ is correct ?

Yes.

 

[Buffu]

Yes.

But now how I am going to evaluate this ? I am confused a bit,

Should I do $\displaystyle \int E_x dx + E_y dy + E_z dz = \int E_z dz = \phi_z$ ?

 

[haruspex]

But now how I am going to evaluate this ? I am confused a bit,

Should I do $\displaystyle \int E_x dx + E_y dy + E_z dz = \int E_z dz = \phi_z$ ?

Your notation confuses me, and it might be confusing you too.
E_x etc. usually mean x component of the field at a general point, i.e. E_x(x,y,z). But in some of your equations you have used E_x to mean E_x(x,0,0) and E_z to mean E(0,0,z).
If you take the potential at the origin to be 0 then $\phi(x,y,z)=\int_SE_x dx + E_y dy + E_z dz$ where S is any path from the origin to (x,y,z). In particular, you could take the path (0,0,0)-(x,0,0)-(x,0,z) to reach (x,0,z), so
$\phi(x,0,z)=\int_0^x E_x(u,0,0) du + \int_0^z E_z(x,0,w) dw$.
But first you need an expression for E_z(x,0,z).

An alternative approach is to consider the potential at (x,0,z) due to some element of the line of charge and integrate along the charge.

 

[Buffu]

Your notation confuses me, and it might be confusing you too.
E_x etc. usually mean x component of the field at a general point, i.e. E_x(x,y,z). But in some of your equations you have used E_x to mean E_x(x,0,0) and E_z to mean E(0,0,z).
If you take the potential at the origin to be 0 then $\phi(x,y,z)=\int_SE_x dx + E_y dy + E_z dz$ where S is any path from the origin to (x,y,z). In particular, you could take the path (0,0,0)-(x,0,0)-(x,0,z) to reach (x,0,z), so
$\phi(x,0,z)=\int_0^x E_x(u,0,0) du + \int_0^z E_z(x,0,w) dw$.
But first you need an expression for E_z(x,0,z).

An alternative approach is to consider the potential at (x,0,z) due to some element of the line of charge and integrate along the charge.

Lol , you are correct, I confused myself with my notation. I was adding potetial compoenent wise, what an idiot.