# Please verify my solution (Electric field cylinder)

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1. Nov 21, 2014

1. The problem statement, all variables and given/known data

Finding the electric field outside of a uniformly charged solid cylinder, of length L and radius R, at any point of its axis.

2. Relevant equations

$$\displaystyle{ \vec{E} = \frac{1}{4 \pi \varepsilon_0} \int \rho(r') \frac{(\vec{r} - \vec{r'})}{\left| \vec{r} - \vec{r'} \right|^3} d^3 r' }$$

$$\rho(r') = \rho_0$$

$$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$$

$$\vec{r'} = x'\hat{i} + y'\hat{j} + z'\hat{k}$$

$$d^3 r' = r' dr' d\theta' dz'$$

$$x' = r' cos \theta'$$

$$y' = r' sin \theta'$$

3. The attempt at a solution

Solving for a point P(0,0,z) outside of the cylinder, thus z > L/2 (assuming half of the cylinder is on the positive z-axis and the other half at the negative z-axis)

$$\displaystyle{ \vec{E} = \frac{1}{4 \pi \varepsilon_0} \rho_0 \int \frac{ (-x'\hat{i} - y'\hat{j} + (z-z')\hat{k} ) }{ (x'^2 + y'^2 + (z-z')^2 )^\frac{3}{2} } d^3 r' }$$

Since it's obvious from the symmetry of the problem that the electric field must be parallel to the z-axis (as the cosine and sine in the integrals will yield 0), then

$$\displaystyle{ \vec{E} = \frac{1}{4 \pi \varepsilon_0} \rho_0 \int \frac{ (z-z')\hat{k} }{ (r'^2 + (z-z')^2 )^\frac{3}{2} } r' dr' d\theta' dz' }$$

$$\displaystyle{ \vec{E} = \frac{1}{4 \pi \varepsilon_0} \rho_0 \hat{k} \int_{\frac{-L}{2}}^{\frac{L}{2}} \left( \int_{0}^{R} \frac{ (z-z')r' dr' }{ (r'^2 + (z-z')^2 )^\frac{3}{2} } \right) dz' \int_{0}^{2\pi} d\theta' }$$

Using u = r'² + (z-z')² and r'dr' = (1/2) du and evaluating the limits, the integral between parentheses will yield

$$\displaystyle{ \vec{E} = \frac{2\pi}{4 \pi \varepsilon_0} \rho_0 \hat{k} \int_{\frac{-L}{2}}^{\frac{L}{2}} \left( \frac{(z-z')}{|z-z'|} - \frac{(z-z')}{\sqrt{R^2 + (z-z')^2 } } \right) dz' }$$

As aforementioned, since P(0,0,z) is outside the cylinder, z > z' and thus the absolute value will be positive then

$$\displaystyle{ \vec{E} = \frac{2\pi}{4 \pi \varepsilon_0} \rho_0 \hat{k} \int_{\frac{-L}{2}}^{\frac{L}{2}} \left( 1 - \frac{(z-z')}{\sqrt{R^2 + (z-z')^2 } } \right) dz' }$$

Using, for the first integral, the formula

$$\displaystyle{ \int \frac{dx}{\sqrt{a^2 + x^2}} = ln (\sqrt{a^2 + x^2} + x) + C }$$

And for the second integral the substitution u = R² + (z-z')², (1/2) du = (z-z') dz', finally leads to

$$\vec{E} = \frac{\rho_0}{2 \varepsilon_0} \left( ln \left( \frac{\sqrt{R^2 + (z+\frac{L}{2})^2 } + (z+\frac{L}{2} )}{\sqrt{R^2 + (z-\frac{L}{2})^2 } + (z-\frac{L}{2} )} \right ) +\sqrt{R^2 + (z-\frac{L}{2})^2} - \sqrt{R^2 + (z+\frac{L}{2})^2 } \right) \hat{k}$$

Is this correct?

2. Nov 22, 2014

### haruspex

I'm not following. The 'first' integral appears to be $$\displaystyle{ \int_{\frac{-L}{2}}^{\frac{L}{2}} \left( 1 \right) dz' }$$.
Certainly the log term in the final equation is dimensionally wrong. Each term should have dimension of length.
A useful check is to see what happens as z tends to infinity. Your integral correctly produces $\frac{R^2L}{z^2}$.

3. Nov 22, 2014

Oops! You're right, I got confused with the first integral, but is the second integral correct, because I'm not getting $\frac{R^2L}{z^2}$

$$\displaystyle{ \vec{E} = \frac{2\pi}{4 \pi \varepsilon_0} \rho_0 \hat{k} \int_{\frac{-L}{2}}^{\frac{L}{2}} dz' - \frac{2\pi}{4 \pi \varepsilon_0} \rho_0 \hat{k} \int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{(z-z')}{\sqrt{R^2 + (z-z')^2 } } dz' }$$

$$u = R^2 + (z-z')^2, du = -2(z-z')dz'$$

$$\displaystyle{ \vec{E} = \frac{2\pi L}{4 \pi \varepsilon_0} \rho_0 \hat{k} + \frac{2\pi}{4 \pi \varepsilon_0} \rho_0 \hat{k} \int_{R^2 + (z+\frac{L}{2})^2}^{R^2 + (z-\frac{L}{2})^2} \frac{u^{-\frac{1}{2}}}{2du} }$$

$$\displaystyle{ \vec{E} = \frac{2\pi }{4 \pi \varepsilon_0} \rho_0 \hat{k} \left( L + \sqrt{R^2 + \left( z - \frac{L}{2}\right)^2 } - \sqrt{R^2 + \left( z + \frac{L}{2}\right)^2 } \right) }$$

4. Nov 22, 2014

### haruspex

I didn't say you should get that as the answer. I said that should be the asymptotic form as z tends to infinity.
It's a non-trivial exercise to check that, but it looks to me that it comes out right.

5. Nov 22, 2014