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\begin{document}

\title{Soluci\selectlanguage{ngerman}ón De Problemas Tema V\selectlanguage{spanish}}



\author[1]{Marian Fernanda Reyes-Carlos }%
\affil[1]{Instituto Tecnológico Superior Zacatecas Occidente}%


\vspace{-1em}



  
  \date{May 08, 2024}


\begingroup
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\maketitle
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\selectlanguage{spanish}
\begin{abstract}
Realizare~ los problemas y sus respectivas soluciones correspondientes a
la unidad \#5%
\end{abstract}%



\sloppy


\subsection*{PROBLEMA 1~}

{\label{672922}}\selectlanguage{spanish}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Problema-1-cargas-lineales/Problema-1-cargas-lineales}
\caption{{This is a caption
{\label{837075}}%
}}
\end{center}
\end{figure}

Tres Cargas En L\selectlanguage{ngerman}ínea

Tres particulas cargadas estan acomodas en linea recta (Fig 1). Calcule
la fuerza neta electrostatica sobre la particula 3.

\par\null\selectlanguage{spanish}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.28\columnwidth]{figures/Q3/Q3}
\caption{{This is a caption
{\label{623630}}%
}}
\end{center}
\end{figure}

\par\null\par\null\par\null\par\null\par\null\par\null

\selectlanguage{ngerman}\subsection*{Solución problema 1~}

{\label{240795}}

En la figura 2 se describen las fuerzas que estan actuando sobre la
carga 3, ahora procedemos a calcular las fuerzas~Q2Q3~y Q3Q1

\par\null



\begin{equation}
\vec{F_{32}}=9*10^9 \frac{Nm^2}{c^2}*\frac {(-4\mu C)(3\mu C)}{(0.20)^2}=-2.7N \hat i
\end{equation}



\begin{equation}
\vec{F_{31}}=9*10^9 \frac{Nm^2}{c^2} *\frac {(-4\mu C)(-8\mu C)}{(0.50m)^2}=1.152N \hat i
\end{equation}



\begin{equation}
\vec{F}=\vec{F_{32}}+\vec{F_{31}}=-1.548
\end{equation}

\subsection*{PROBLEMA 2~}

{\label{579061}}\par\null\par\null\par\null\par\null\selectlanguage{spanish}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Fuerza-electrostatica/Fuerza-electrostatica}
\caption{{This is a caption
{\label{909986}}%
}}
\end{center}
\end{figure}

\par\null\par\null\par\null\par\null

Calcular la fuerza electrostatica neta sobre la carga~Q3~debido a las
cargas Q1 y Q2~sin tomar en cuenta la direccion en la que esten sus
fuerzas.

\selectlanguage{ngerman}\subsection*{SOLUCIÓN PROBLEMA 2~}

{\label{799842}}\par\null

\begin{equation}
\vec{F_{31}}=9*10^9 \frac{Nm^2}{c^2} *\frac {(65\mu C)(-86\mu C)}{(0.6m)^2}=139.75N
\end{equation}

\begin{equation}
\vec{F_{31}}=9*10^9 \frac{Nm^2}{c^2} *\frac {(65\mu C)(50\mu C)}{(0.3m)^2}=325N
\end{equation}

Ahora que tenemos las magnitudes utilizaremos trigonometria para
calcular la direccion y obtener la magnitud neta.

\par\null

\begin{equation}
\vec{F_{31}}=\vec F_{31x} \hat i + \vec F_{31y}\hat j
\end{equation}



\begin{equation}
\vec{F_{31x}}=\vec F_{31}Cos30=+121.02
\end{equation}



\begin{equation}
\vec{F_{31y}}=\vec F_{31}Sin30=-69.87
\end{equation}



\begin{equation}
\vec{F_{32}}=\vec F_{32x}\hat i + \vec F_{32y}\hat j
\end{equation}



\begin{equation}
\vec{F_{32x}}=0
\end{equation}



\begin{equation}
\vec{F_{32y}}=\vec F_{32}
\end{equation}



\begin{equation}
\vec{F_{32}}=325N \hat j
\end{equation}

Ahora contamos ya con los valores procederemos a sustituir

\par\null

\begin{equation}
\vec{F}=\vec F_{31}+ \vec F_{32}
\end{equation}



\begin{equation}
\vec{F}=(121.02N)\hat i - (69.87N)\hat j + (325N)\hat j
\end{equation}

Agrupamos~

\par\null

\begin{equation}
\vec{F}=121.02N\hat i + 255.13 N\hat j
\end{equation}

Aplicamos el teorema de pitagoras para conocer la magnitud del la fuerza

\begin{equation}
\vec{|F|}=\sqrt{(121.02)^2 + (255.13)^2}=282.38N
\end{equation}

Ahora calculamos el angulo

\begin{equation}
{\theta}=Tan^{-1} * (\frac {255.13}{121.01})
\end{equation}

Resolvemos para obtener reslutado

\par\null

\begin{equation}
\theta=64.62^\circ
\end{equation}

\selectlanguage{spanish}
\FloatBarrier
\end{document}

