- #1
YK0001
- 3
- 1
- Homework Statement
- A uniformly charged disk with the total charge ##Q##, has an exterior radius of ##b_{1}## and the disc is an annular disc with the interior radius of ##b_{0}##.
A test charge ##q_{t}## has been placed ##\overrightarrow{A}\text{meters}## away from the center of the disk ##\overrightarrow{O}##.
$$\overrightarrow{O} =\begin{bmatrix}0\\0\end{bmatrix}\text{ and }\overrightarrow{A} =\begin{bmatrix}x\\y\end{bmatrix}$$
Calculate the electrostatic force (scalar magnitude with direction or a vector quantity) on ##q_{t}## caused by the disk.
- Relevant Equations
- Coulomb's law: ##\overrightarrow{F}_{12} = \frac{kq_{1}q_{2}}{\overrightarrow{r_{12}}^{2}}##
Volume by revolving area bound by a function: ##A(x) = \pi(F(x)^2)\text{ and }V = \int_{a}^{b}A(x)dx\text{ or } Y = \int_{c}^{d}A(y)dy##
Since we know that ##Q = \text{Total charge uniformly distributed in the disc}## and the area of the disc is just ##\pi((b_1-b_0)^2)##.
We can define ##dQ_{\tiny{|}}## as the charge of an infinitesimally small point on the disk, and ##Q_{\tiny{|}}## as the chage of an infinitesimally small cross sectional line from the disc, such that ##Q = \pi(\int dQ_{\tiny{|}})^2##. We can then calculate the electrostatic force on ##q_t## from ##dQ_{\tiny{|}}## to be ##dF_{\tiny{|}} = \frac{kQ_{\tiny{|}}q_t}{(r)^2} = \frac{kQ_{\tiny{|}}q_t}{x^2+b^2-2xbcos(\theta)}##, then we can conclude that: $$F_{\tiny{|}} = \int\frac{kdQ_{\tiny{|}}q_{t}}{x^2+b^2-2xbcos(\theta)} = kq_{t}\int\frac{dQ_{\tiny{|}}}{x^2+b^2-2xbcos(\theta)}$$.
This is where the problem arises, I don't know what to integrate the right side to respect with, I've hypothesized that it's with respect to ##Q_{\tiny{|}}## so it looks like ##\int [...] dQ_{\tiny{|}}## because that is the infinitesimally small term already present on that side of the equation, but I am unsure if this is the correct assumption to make, I'm also thinking that there are 3 changing variables in that integral (x, b, theta (theta not really, since we can get that in terms of x and b), but I'm not sure what that integral needs to be).
I have drawn a diagram showing the disc and have attached that as well, the blue shaded areas are charged.
We can define ##dQ_{\tiny{|}}## as the charge of an infinitesimally small point on the disk, and ##Q_{\tiny{|}}## as the chage of an infinitesimally small cross sectional line from the disc, such that ##Q = \pi(\int dQ_{\tiny{|}})^2##. We can then calculate the electrostatic force on ##q_t## from ##dQ_{\tiny{|}}## to be ##dF_{\tiny{|}} = \frac{kQ_{\tiny{|}}q_t}{(r)^2} = \frac{kQ_{\tiny{|}}q_t}{x^2+b^2-2xbcos(\theta)}##, then we can conclude that: $$F_{\tiny{|}} = \int\frac{kdQ_{\tiny{|}}q_{t}}{x^2+b^2-2xbcos(\theta)} = kq_{t}\int\frac{dQ_{\tiny{|}}}{x^2+b^2-2xbcos(\theta)}$$.
This is where the problem arises, I don't know what to integrate the right side to respect with, I've hypothesized that it's with respect to ##Q_{\tiny{|}}## so it looks like ##\int [...] dQ_{\tiny{|}}## because that is the infinitesimally small term already present on that side of the equation, but I am unsure if this is the correct assumption to make, I'm also thinking that there are 3 changing variables in that integral (x, b, theta (theta not really, since we can get that in terms of x and b), but I'm not sure what that integral needs to be).
I have drawn a diagram showing the disc and have attached that as well, the blue shaded areas are charged.