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.)