Potential due to a line charge

In summary: Yes I know how to evaluate integral, I will do that and let you know if I have any problem. Thank you so much for guiding me. I am really grateful to you. :)Lol :H:H , you are correct, I confused myself with my notation. I was adding potetial compoenent wise, what an idiot. Yes I know how to evaluate integral, I will do that and let you know if I have any problem. Thank you so much for guiding me. I am really grateful to you. :)You're welcome.You're welcome.
  • #1
Buffu
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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,

Physics potential.png


##\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 ?
 
Last edited:
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  • #2
Buffu said:
##\displaystyle \phi (x,0,z) =\phi_x + \phi_z ##
Why?
Your φx is shorthand for φ(x,0,0). Similarly φz.
 
  • #3
haruspex said:
Why?
Your φx is shorthand for φ(x,0,0). Similarly φz.

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 ?
 
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  • #4
Buffu said:
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.)
 
Last edited:
  • #5
haruspex said:
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 ?
 
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  • #6
Buffu said:
So ##\displaystyle \phi(x, 0, z) = \phi_x + \int_{(x, 0 , 0 )}^{(x,0,z)} \vec E d\vec s## is correct ?
Yes.
 
  • #7
haruspex said:
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## ?
 
  • #8
Buffu said:
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.
Ex etc. usually mean x component of the field at a general point, i.e. Ex(x,y,z). But in some of your equations you have used Ex to mean Ex(x,0,0) and Ez 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 Ez(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.
 
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  • #9
haruspex said:
Your notation confuses me, and it might be confusing you too.
Ex etc. usually mean x component of the field at a general point, i.e. Ex(x,y,z). But in some of your equations you have used Ex to mean Ex(x,0,0) and Ez 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 Ez(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 :H:H , you are correct, I confused myself with my notation. I was adding potetial compoenent wise, what an idiot.
 

Related to Potential due to a line charge

1. What is the formula for calculating the potential due to a line charge?

The formula for calculating the potential due to a line charge is V = kλ/2πε0r, where V is the potential, k is the Coulomb's constant, λ is the line charge density, ε0 is the permittivity of free space, and r is the distance from the line charge.

2. How does the potential due to a line charge vary with distance?

The potential due to a line charge varies inversely with distance. This means that as the distance from the line charge increases, the potential decreases.

3. What is the direction of the electric field due to a line charge?

The electric field due to a line charge is always directed radially outward or inward from the line charge, depending on the sign of the charge.

4. Can the potential due to a line charge ever be negative?

Yes, the potential due to a line charge can be negative if the line charge is negative. This means that the potential at a certain point is lower than the potential at infinity.

5. How does the potential due to a line charge compare to the potential due to a point charge?

The potential due to a line charge is different from the potential due to a point charge. The potential due to a line charge decreases logarithmically with distance, while the potential due to a point charge decreases inversely with distance. This means that the potential due to a line charge decreases more slowly with distance compared to a point charge.

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