Voltage change away from long line of uniform charge

[Pyuruku]

Homework Statement

Find the voltage change going from 5.0cm to 10.0cm along the radius away from a long line of uniform charge density 5.0 μC/m

Homework Equations

V = (kq) / r (I believe... I'm very lost)

The Attempt at a Solution

ΔV = ∫.05 to .1 2∏rLdL = 2∏r((L^2) / 2 | from .05 to .1) = 2∏r ( .1^2 / 2 - .05^2 / 2) = 0.0235r

Obviously I'm doing this wrong... I'm struggling with the topic of electricity so I apologize if this is too elementary...

 

[gneill]

Homework Statement

Find the voltage change going from 5.0cm to 10.0cm along the radius away from a long line of uniform charge density 5.0 μC/m

Homework Equations

V = (kq) / r (I believe... I'm very lost)

The Attempt at a Solution

ΔV = ∫.05 to .1 2∏rLdL = 2∏r((L^2) / 2 | from .05 to .1) = 2∏r ( .1^2 / 2 - .05^2 / 2) = 0.0235r

Obviously I'm doing this wrong... I'm struggling with the topic of electricity so I apologize if this is too elementary...

Hi Pyuruku, Welcome to Physics Forums.

Why not start with the expression for the electric field strength at a distance r from an infinite line of charge? Then integrate along the path between the two radial distances.

 

[Pyuruku]

Would that be E = \frac{\lambda}{2\pi r\epsilon}?

If so...

\frac{\lambda}{2\pi\epsilon}\int_{0.05}^{0.1} \! \frac{dr}{r} = \frac{\lambda}{2\pi\epsilon} ( ln(r) \bigr|_{0.05}^{0.1}) = \frac{\lambda}{2\pi\epsilon}ln(\frac{0.1}{0.05}) = \frac{5.0}{2\pi(8.85x10^{-12}}ln(\frac{0.1}{0.05}) = 6.23x10^{10}

Does that seem correct? I can't check the answer for myself (practice worksheet)

 

[gneill]

Would that be E = \frac{\lambda}{2\pi r\epsilon}?

If so...

\frac{\lambda}{2\pi\epsilon}\int_{0.05}^{0.1} \! \frac{dr}{r} = \frac{\lambda}{2\pi\epsilon} ( ln(r) \bigr|_{0.05}^{0.1}) = \frac{\lambda}{2\pi\epsilon}ln(\frac{0.1}{0.05}) = \frac{5.0}{2\pi(8.85x10^{-12}}ln(\frac{0.1}{0.05}) = 6.23x10^{10}

Does that seem correct? I can't check the answer for myself (practice worksheet)

Fine, except that the charge density is not 5 coulombs per meter, it's 5 microcoulombs per meter. So your result is off by only six orders of magnitude

 

[Pyuruku]

\frac{\lambda}{2\pi\epsilon}\int_{0.05}^{0.1} \! \frac{dr}{r} = \frac{\lambda}{2\pi\epsilon} ( ln(r) \bigr|_{0.05}^{0.1}) = \frac{\lambda}{2\pi\epsilon}ln(\frac{0.1}{0.05}) = \frac{5.0x10^{-6}}{2\pi(8.85x10^{-12})}ln(\frac{0.1}{0.05}) = 6.232 x 10^{4} V

(I believe this is now correct)

 

[gneill]

Yup, looks okay now. Be sure to make your final result match the given data in terms of significant figures.