Firstly let's assume we have a charged stick$^1$ in a 2-d vector space like the following:
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.
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.
Force of one charge on another in a 3-d vector space
physicsvector analysis
Related Solutions
I suggest you start by setting the problem up just like in the picture I have attached.
ASSUMING ALL CHARGES ARE POSITIVE AS IN THE PROBLEM STATEMENT
You can calculate each of the forces separately:
$$ F_1 = k\frac{q1q3}{r_{13}^2} = k\frac{(2)(3)}{8^2}\hat{i} $$
$F_2$ is slightly more complicated
$$ F_2 = k\frac{q2q3}{r_{23}^2} = k\frac{(2)(3)}{(\sqrt{8^2+5^2})^2} $$
Now you have to break down $F_2$ into two different vectors $F_{2x}$ and $F_{2y}$, which are the projections of $F_2$ on the x,y axis.
You can do this by figuring out the angle $\theta$ between $F_1$ and $F_2$.
Doing some simple trig:
$$ \theta = \arctan\left(\frac{5}{8}\right) $$
Then you can say:
$$ F_{2x} = F_2\cos(\theta)\hat{i} $$
$$ F_{2y} = F_2\cos(\theta)\hat{j} $$
Now that all your forces are represented as vectors on the $x-y$ plane, you can find the resultant by just adding the vectors.
Can you proceed from here?
We begin with the expression for the electric field
$$\vec E(\vec r')=k_e\int_{V}\frac{\rho(\vec r_s)(\vec r'-\vec r_s)}{|\vec r'-\vec r_s|^3}\,dV_s\tag1$$
Now, if we integrate $(1)$ from $\vec r_1$ to $\vec r$, and let $|\vec r_1|\to \infty$, we find that
$$\begin{align} -\lim_{|\vec r_1|\to \infty}\int_{\vec r_1}^{\vec r}\vec E(\vec r')\cdot d\vec r'&=-\lim_{|\vec r_1|\to \infty}\int_{\vec r_1}^{\vec r}\left(k_e\int_{V}\frac{\rho(\vec r_s)(\vec r'-\vec r_s)}{|\vec r'-\vec r_s|^3}\,dV_s\right)\cdot d\vec r'\\\\ &=-\lim_{|\vec r_1|\to \infty}\int_{\vec r_1}^{\vec r}\left(-k_e\int_{V}\rho(\vec r_s)\nabla '\frac{1}{|\vec r'-\vec r_s|}\,dV_s\right)\cdot d\vec r'\\\\ &=k_e \lim_{|\vec r_1|\to \infty}\int_{V} \rho(\vec r_s)\int_{\vec r_1}^{\vec r}\nabla '\frac{1}{|\vec r'-\vec r_s|}\cdot d\vec r'\,dV_s\\\\ &=k_e \lim_{|\vec r_1|\to \infty}\int_{V} \rho(\vec r_s)\left(\frac1{|\vec r-\vec r_s|}-\frac1{|\vec r_1-\vec r_s|}\right)\,dV_s\\\\ &=k_e \int_{V} \frac{\rho(\vec r_s)}{|\vec r-\vec r_s|}\,dV_s \end{align}$$
as was to be shown!
Best Answer
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 $2\pi$