Ley de Coulomb (Ejercicio Resuelto 009)
Calcular la fuerza total que ejerce el disco cargado uniformemente de radio \(R\) sobre una carga positiva \(q\) que se encuentra en el punto \(P\). La distanacia entre el centro del disco y la carga es \( 2\,R \).
SOLUCIÓN.
Datos. \(\qquad R\), \(\qquad q\), \(\qquad F=\,?\)
\[ \begin{aligned} &dF_z=dF\cos\theta\\ &dF=\frac{1}{4\pi\varepsilon_o}\frac{q\,dQ}{d^{2}}\;;\;\sigma=\frac{dQ}{dA}\Rightarrow\,dQ=\sigma dA=\sigma 2\pi rdr\\ &dF=\frac{1}{4\pi\varepsilon_o}\frac{q\sigma 2\pi dr}{d^{2}}\;;\;\cos\theta=\frac{2R}{d}\\ &dFz=\frac{1}{4\pi\varepsilon_o}\frac{q\sigma 2\pi rdr}{d^{2}}\frac{2R}{d}\;;\;d=\sqrt{r^{2}+\left(2R\right)^{2}}\\ &dF_z=\frac{1}{4\pi\varepsilon_o}\frac{q\sigma2\pi rdr\,2R}{\left(r^{2}+4R^{2}\right)^{3/2}}\\ &F_z=\int_{0}^{R}\frac{1}{4\pi\varepsilon_o}\frac{q\sigma2\pi rdr\,2R}{\left(r^{2}+4R^{2}\right)^{3/2}}\\ &F_z=\frac{1}{4\pi\varepsilon_o}q\sigma2\pi\,2R\int_{0}^{R}\frac{r}{\left(r^{2}+4R^{2}\right)^{3/2}}dr\\ &F_z=\frac{1}{4\pi\varepsilon_o}q\sigma2\pi2R\left[-\frac{1}{\sqrt{r^{2}+4R^{2}}}\right]_{0}^{R}\\ &F_z=\frac{1}{4\pi\varepsilon_o}q\sigma2\pi2R\left(-\frac{1}{\sqrt{R^{2}+4R^{2}}}+\frac{1}{2R}\right)\\ &F_z=\frac{1}{4\pi\varepsilon_o}q\sigma2\pi2R\left(\frac{1}{2R}-\frac{1}{R\sqrt{5}}\right)\\ &F_z=\frac{1}{4\pi\varepsilon_o}q\sigma2\pi2R\left(\frac{1}{R}\left(\frac{1}{2}-\frac{1}{\sqrt{5}}\right)\right)\\ &F_z=\frac{1}{4\pi\varepsilon_o}q\sigma4\pi\left(\frac{1}{2}-\frac{1}{\sqrt{5}}\right)\;;\;\sigma=\frac{Q}{\pi R^{2}}\\ &F_z=\frac{1}{4\pi\varepsilon_o}\frac{q\,Q4\pi}{\pi R^{2}}\left(\frac{1}{2}-\frac{1}{\sqrt{5}}\right)\\ &\boxed{F_z=\frac{1}{\pi\varepsilon_o}\frac{qQ}{ R^{2}}\left(\frac{1}{2}-\frac{1}{\sqrt{5}}\right)} \end{aligned} \]
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