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

• KillerZ
In summary: Ok:Ey = -\frac{\frac{Q}{L}}{4\pi\epsilon_{0}}\left[\frac{1}{\sqrt{x^{2}}} - \frac{1}{\sqrt{L^{2} + x^{2}}}\right]Excellent.In summary, we are finding the electric field E at distance x from the end of a thin rod of length L with total charge Q. We have derived an expression for E in component form, using the formula E = \frac{1}{4\pi\epsilon_{0}}\frac{\Delta Q}{r\sqrt{r^{2} + (\frac{L}{2})^{2}}}, and integrating to find the x and
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.

KillerZ said:
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.

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

(Ei)x = Eicos$$\theta$$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.

Set up the integral and derive your own formula.

Ok here is what I got for the x component:

(Ei)x = Eicos$$\theta$$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 = $$\frac{x}{\sqrt{y_{i}^{2} + x^{2}}}$$

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

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

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

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

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

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

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

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

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

Excellent!

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

(Ei)y = Eisin$$\theta$$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 = $$\frac{y_{i}}{\sqrt{y_{i}^{2} + x^{2}}}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## 1. How do you calculate the electric field of a thin rod?

To calculate the electric field of a thin rod of length L with total charge Q, you can use the equation E = kQ/L, where k is the Coulomb's constant equal to 8.99 x 10^9 Nm^2/C^2. This equation assumes that the rod is infinitely thin and has a uniform charge distribution along its length.

## 2. What is the direction of the electric field around a thin rod?

The electric field around a thin rod is radial, meaning it points away from the rod if the charge is positive and towards the rod if the charge is negative. The electric field also follows the right-hand rule, where if you point your thumb in the direction of the electric field, your fingers will curl in the direction of the electric field lines.

## 3. Can the electric field of a thin rod be negative?

Yes, the electric field of a thin rod can be negative if the total charge Q is negative. This will cause the electric field lines to point towards the rod instead of away from it.

## 4. How does the electric field of a thin rod change with distance?

The electric field of a thin rod follows the inverse square law, which means that as you move further away from the rod, the electric field strength decreases proportionally to the square of the distance. This means that the electric field will be strongest closest to the rod and will decrease as you move further away.

## 5. Can the electric field of a thin rod be affected by external charges?

Yes, the electric field of a thin rod can be affected by external charges. If there are other charges in the vicinity of the rod, their electric fields will interact and change the overall electric field around the rod. In some cases, this could result in a non-uniform electric field around the rod.

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