Calculating Electric Field of a Thin Rod of Length L with Total Charge Q

[KillerZ]

Homework Statement

A thin rod of length L with total charge Q. Find an expression for the electric field E at distance x from the end of the rod. Give your answer in component form.

Homework Equations

Well this isn't a infinite line of charge so this formula should work:

E = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r\sqrt{r^{2} + (\frac{L}{2})^{2}}}

The Attempt at a Solution

I think I will have to integrate because I cannot cancel components because there is no symmetry in the diagram all electric fields will point in the positive x, negative y direction.

 

[Doc Al]

Well this isn't a infinite line of charge so this formula should work:

E = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r\sqrt{r^{2} + (\frac{L}{2})^{2}}}

Where did this formula come from?

Yes, you'll have to integrate. Hint: Break the line charge into small elements of length dy.

 

[KillerZ]

I got that formula from and example in my physics book it is the result after integrating this formula:

(E_i)_x = E_icos\theta_i = \frac{1}{4\pi\epsilon_{0}}\frac{\Delta Q}{r_{i}^{2}}cos\theta_i

I think that formula is for a point that bisects the rod only.

 

[Doc Al]

Set up the integral and derive your own formula.

 

[KillerZ]

Ok here is what I got for the x component:

(E_i)_x = E_icos\theta_i = \frac{1}{4\pi\epsilon_{0}}\frac{\Delta Q}{r_{i}^{2}}cos\theta_i

r = \sqrt{y_{i}^{2} + x^{2}}

cos\theta_i = \frac{x}{\sqrt{y_{i}^{2} + x^{2}}}

(E_i)_x = \frac{1}{4\pi\epsilon_{0}}\frac{\Delta Q}{y_{i}^{2} + x^{2}}\frac{x}{\sqrt{y_{i}^{2} + x^{2}}}

= \frac{1}{4\pi\epsilon_{0}}\frac{x\Delta Q}{(y_{i}^{2} + x^{2})^{3/2}}

E_x = \sum^{N}_{i = 1}(E_i)_x = \frac{1}{4\pi\epsilon_{0}}\sum^{N}_{i = 1}\frac{x\Delta Q}{(y_{i}^{2} + x^{2})^{3/2}}

\Delta Q = \lambda\Delta y = (\frac{Q}{L})\Delta y

E_x = \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\sum^{N}_{i = 1}\frac{x\Delta y}{(y_{i}^{2} + x^{2})^{3/2}}

= \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\int_{0}^{L}\frac{xdy}{(y^{2} + x^{2})^{3/2}}

= \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\frac{y}{x\sqrt{y^{2} + x^{2}}}\left|_{0}^{L}

= \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\left[\frac{L}{x\sqrt{L^{2} + x^{2}}} - \frac{0}{x\sqrt{0^{2} + x^{2}}}\right]

= \frac{1}{4\pi\epsilon_{0}}\frac{Q}{x\sqrt{L^{2} + x^{2}}}

 

[Doc Al]

Excellent!

 

[KillerZ]

Ok here is what I got for the y component, I am not sure if this one is right:

(E_i)_y = E_isin\theta_i = \frac{1}{4\pi\epsilon_{0}}\frac{\Delta Q}{r_{i}^{2}}sin\theta_i

r = \sqrt{y_{i}^{2} + x^{2}}

sin\theta_i = \frac{y_{i}}{\sqrt{y_{i}^{2} + x^{2}}}

(E_i)_y = \frac{1}{4\pi\epsilon_{0}}\frac{\Delta Q}{y_{i}^{2} + x^{2}}\frac{y_{i}}{\sqrt{y_{i}^{2} + x^{2}}}

= \frac{1}{4\pi\epsilon_{0}}\frac{y_{i}\Delta Q}{(y_{i}^{2} + x^{2})^{3/2}}

E_y = \sum^{N}_{i = 1}(E_i)_y = \frac{1}{4\pi\epsilon_{0}}\sum^{N}_{i = 1}\frac{y_{i}\Delta Q}{(y_{i}^{2} + x^{2})^{3/2}}

\Delta Q = \lambda\Delta y = (\frac{Q}{L})\Delta y

E_y = \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\sum^{N}_{i = 1}\frac{y_{i}\Delta y}{(y_{i}^{2} + x^{2})^{3/2}}

= \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\int_{0}^{L}\frac{ydy}{(y^{2} + x^{2})^{3/2}}

= \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\frac{y}{y\sqrt{y^{2} + x^{2}}}\left|_{0}^{L}

= \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\left[\frac{L}{L\sqrt{L^{2} + x^{2}}} - \frac{0}{0\sqrt{0^{2} + x^{2}}}\right]

= \frac{1}{4\pi\epsilon_{0}}\frac{Q}{L\sqrt{L^{2} + x^{2}}}

 

[Doc Al]

= \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\int_{0}^{L}\frac{ydy}{(y^{2} + x^{2})^{3/2}}

Good.

= \frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\frac{y}{y\sqrt{y^{2} + x^{2}}}\left|_{0}^{L}

Cancel those y's. (And be careful with the sign of the integral.)

 

[KillerZ]

Ok, I made the changes:

= -\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\frac{1}{\sqrt{y^{2} + x^{2}}}\left|_{0}^{L}

= -\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\left[\frac{1}{\sqrt{L^{2} + x^{2}}} - \frac{1}{\sqrt{0^{2} + x^{2}}}\right]

= -\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\left[\frac{1}{\sqrt{L^{2} + x^{2}}} - \frac{1}{\sqrt{x^{2}}}\right]

 

[Doc Al]

That looks good. Move that outside minus sign inside. (Switch the order of subtraction in the brackets.)