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Firstly let's assume we have a charged stick$^1$ in a 2-d vector space like the following:
enter image description here
On the x-axis we have a charge $q_0$ in a distance $d$. The stick has a length $l$ and a total charge $Q_C$. Now we want to calculate the force of the charged stick on $q_0$.
The basic formula for this would be $$\vec{F} = \frac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} \cdot \hat{r}$$
$\epsilon_0$: permittivity of free space (constant)
$q_1$: first charge ($q_0$ in this case)
$q_2$: second charge (the charge of the stick)
$r^2$: distance between the two charges raised to the second power $(|\vec{r}|^2)$
$\hat{r}$: unit vector $\frac{\vec{r}}{|\vec{r}|}$
To tackle this problem we determine a part of the stick as $d_y$ which has the coordinates $(y; 0)$ $$\vec{r} = d\hat{x}-y \hat{y}$$ $$|\vec{r}| = \sqrt{d^2 + y^2}$$ $$|\vec{r}|^2 = d^2 + y^2$$ $$\hat{r} = \frac{d}{(d^2 + y^2)^{1/2}}\cdot \hat{x} - \frac{y}{(d^2+y^2)^{1/2}}\cdot \hat{y}$$ This leads us to: $$\frac{1}{4\cdot \pi \cdot \epsilon_0} \cdot \frac{q_0 \cdot \lambda \cdot dy}{(d^2+y^2)}\cdot \bigg[\frac{d}{(d^2+y^2)^{1/2}} \cdot \hat{x} - \frac{y}{(d^2+y^2)^{1/2}} \cdot \hat{y} \bigg]$$ $$=\frac{1}{4\cdot \pi \cdot \epsilon_0} \cdot q_0 \cdot \lambda \cdot \bigg[\frac{d}{(d^2+y^2)^{3/2}} \cdot \hat{x} - \frac{y}{(d^2+y^2)^{3/2}} \cdot \hat{y} \bigg] \cdot dy$$ $$\int^{l/2}_{-l/2}{\vec{F}} =\frac{1}{4\cdot \pi \cdot \epsilon_0} \cdot q_0 \cdot \lambda \cdot \bigg[\int^{l/2}_{-l/2}{\frac{d}{(d^2+y^2)^{3/2}} \cdot \hat{x} - \frac{y}{(d^2+y^2)^{3/2}} \cdot \hat{y}} \bigg] \cdot dy$$
Because of the symmetry the y-components neutralize each other. $$\int^{l/2}_{-l/2}{\vec{F}} =\frac{1}{4\cdot \pi \cdot \epsilon_0} \cdot q_0 \cdot \lambda \cdot d \bigg[\int^{l/2}_{-l/2}{\frac{1}{(d^2+y^2)^{3/2}}\cdot dy} \bigg] \cdot \hat{x}$$
Well now we have a charged ring and a second charge in z-direction.
enter image description here
My approach was again to determine a $dy$, find the vectors, etc.
$$\vec{r} = -R\hat{y} + d\hat{z}$$ $$|\vec{r}| = \sqrt{R^2+d^2}$$ $$|\vec{r}|^2 = R^2+d^2$$ $$\hat{r} = \frac{-R\hat{y} + d\hat{z}}{(R^2+d^2)^{1/2}}$$
$$\vec{F} = \frac{1}{4\cdot \pi \cdot \epsilon_0} \cdot \frac{q_0 \cdot \lambda \cdot dy}{R^2+d^2} \cdot \frac{-R \hat{y} + d \hat{z}}{(R^2+d^2)^{1/2}}$$ $$= \frac{1}{4\cdot \pi \cdot \epsilon_0} \cdot q_0 \cdot \lambda \bigg[ \cdot \frac{-R \hat{y}}{(R^2+d^2)^{3/2}} + \frac{d \hat{z}}{(R^2+d^2)^{3/2}} \bigg]\cdot dy$$
I have the feeling that I'm missing something here because I didn't pay attention to the x-axis. Also I wonder how the integral would look like - is it similar to the stick but with the circumference as limits (circle integral)?
$^1$ Didn't have a better translation for that. I am open for suggestions.

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The best way to approach this is to write the Cartesian components of $\vec r$ in terms of cylindrical coordinates. In the horizontal plane a particle with charge $dq=\lambda dl$ is at distance $R$ from $0$, at an angle $\theta$. $$dl=Rd\theta$$ Similarly to your previous example, the $x$ and $y$ components will cancel out. The $z$ component is $$dF_z=\frac{1}{4\pi\epsilon_0}q_0\lambda \frac{d}{(R^2+d^2)^{3/2}}R\ d\theta$$ Now all you need is to integrate from $0$ to $2\pi$

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  • $\begingroup$ and because it's a vector we'd add $\hat{z}$ as unit vector. Is this correct? $\endgroup$ – TimSch Nov 6 '18 at 22:47
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    $\begingroup$ Correct. You add $\hat z$ to the final answer $\endgroup$ – Andrei Nov 6 '18 at 22:50

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