// Numbas version: finer_feedback_settings {"name": "FIS5047_S6_Numbas", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": [""], "variable_overrides": [[]], "questions": [{"name": "Superposici\u00f3n de campos el\u00e9ctricos", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Picture_1_2tek4ZS.png", "/srv/numbas/media/question-resources/Picture_1_2tek4ZS.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Jon Goss", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3925/"}, {"name": "Marcelo Calder\u00f3n", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/17160/"}, {"name": "Ivan Munoz", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18334/"}], "tags": ["coulomb", "elec", "field", "tut1", "Vector", "vector"], "metadata": {"description": "

Evaluación de la superposición vectorial de campos provenientes de cuatro cargas puntuales. Este es un problema de suma de vectores, magnitudes de vectores y productos escalares (puntos) con un poco de trigonometría.

", "licence": "None specified"}, "statement": "

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Un cuadrado de lado {sidecm} $cm$ posee distintas cargas eléctricas situadas en sus esquinas. Las cuales se denotan como $q_1$, $q_2$, $q_3$ y $q_4$ y cuyas magnitudes son respectivamente {q1}, {q2}, {q3} y {q4} $nC$.

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En esta pregunta interactiva se le pide determinar el campo eléctrico total en el centro del cuadrado incluyendo tanto la magnitud como la dirección de este (es un vector).

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Esta pregunta es un ejemplo del principio de superposición, concerniente a la naturaleza física y matemática de un campo vectorial.

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Use el valor de las constanes brindadas en los videos de clases y en sesiones de tutoría, puede usar hasta 4 cifras significativas y asuma que el medio circundante es el vacío, donde $\\epsilon_0=8,8\\times10^{-12}$

", "advice": "

\n

Este es un ejercicio de que involucra la superposición de cuatro campos eléctricos diferentes, lo cual equivale a la suma de cuatro vectores.  Cada campo proviene de la ley de Coulomb (para cargas puntuales), por lo tando se debe considerar el factor

\n

$\\displaystyle {q_i\\over 4\\pi\\varepsilon_0 r^2}$,

\n

donde, debido a que el punto de interés es el centro del cuadrado, la distancia $r$ es la misma para cada una de las cuatro cargas (en este caso {side*100}/$\\sqrt{2}\\approx$ {siground(r*100,4)} cm). En cada caso el campo apunta radialmente alejandose de la respectiva carca puntual tal que para $q_1$ este es hacia abajo e izquierda en igual medida, para $q_2$ este es hacia arriba y hacia la izquierda y ocurriendo algo similar con las demás. Para cargas negativas, el signo $q$ puede ser pensado como una inversión en el sentido del campo eléctrico (los campos se dirigen hacia las cargas negativas). Debe ser cuidadosa o cuidadoso al determinar la dirección y note que los diagramas generalmente ayudan mucho!

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El valor de $E$ puede obtenerse fácilmente y la dirección en cada caso está dada por

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$\\displaystyle{1\\over\\sqrt{2}}\\left(\\pm \\hat{i} \\pm \\hat{j}\\right)$, 

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con las combinaciones de $\\pm$ dependiendo de qué carga estamos considerando y con $\\sqrt{2}$ siendo un factor que proviene del hecho de que es un vector unitario.

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Una vez que se ha obtenido cada campo vectorial, se pueden sumar por componente (aplicar el principio de superposición) dando el campo $E$ total en el centro del cuadrado.

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Finalmente, el ángulo con la línea que une $q_2$ y $q_4$ se puede obtener observando que el producto escalar del campo total con la dirección del desplazamiento de $q_2$ a $q_4$ da un ángulo:

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$\\displaystyle{\\vec{E}.\\left(-\\frac{\\hat{i}}{\\sqrt{2}}+\\frac{\\hat{j}}{\\sqrt{2}}\\right)=\\left|\\vec{E}\\right|\\cos(\\theta)}$

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tal que

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$\\displaystyle{-E_x+E_y=\\sqrt{2}\\left|\\vec{E}\\right|\\cos(\\theta)\\Rightarrow \\theta=\\cos^{-1}\\left(\\frac{E_y-E_x}{\\sqrt{2}\\left|\\vec{E}\\right|}\\right)}$.

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Entonces solo queda determinar el valor de theta que se encuentra en el rango de ángulo requerido.

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$y$-component of the E-field from q1(N/C).

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Point charge in the bottom left corner, nC.

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Relative permittivity of air.

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Magnitud del campo eléctrico en el centro del cuadrado.

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Length of side of square in cm.

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$x$-component of the E-field from q2 (N/C).

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Angle between the line between q2 and q4 and the E-field

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$y$-component of the E-field from q4 (N/C).

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$y$-component of the E-field due to q2 (N/C).

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$x$-component of the E-field from q4 (N/C).

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Point charge in the top left corner, nC.

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Distance from the charges to the centre of the square in m.

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Point charge in bottom right corner, nC.

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$x$-component of the E-field from q1 (N/C).

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Permittivity of free space in F/m to 4 significant figures.

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The constant, $1/4\\pi\\varepsilon$ in units of m/F.

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$y$-component of the E-field from q3 (N/C).

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$x$-component of the E-field from q3 (N/C).

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Point charge in top right corner in nC.

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r^2, cm

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Side length in m.

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q1 in C

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q2 in C

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q3 in C

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q4 in C

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Determine el campo eléctrico en el centro del cuadrado y que produce la carga $q_1$ ({q1}nC).

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$E_x(q_1)=$[[0]]V/m

\n

$E_y(q_1)=$[[1]]V/m

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Determine el campo eléctrico en el centro del cuadrado y que produce la carga $q_2$ ({q2}nC).

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$E_x(q_2)=$[[0]]V/m

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$E_y(q_2)=$[[1]]V/m

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Determine el campo eléctrico en el centro del cuadrado y que produce la carga $q_3$ ({q3}nC).

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$E_x(q_3)=$[[0]]V/m

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$E_y(q_3)=$[[1]]V/m

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Determine el campo eléctrico en el centro del cuadrado y que produce la carga $q_4$ ({q4}nC).

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$E_x(q_4)=$[[0]]V/m

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$E_y(q_4)=$[[1]]V/m

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Determine el campo eléctrico neto (total) en el centro del cudrado debido a la presencia de las cuatro cargas.

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$E_x=$[[0]]V/m

\n

$E_y=$[[1]]V/m

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¿Cuál es el ángulo, $\\theta$, entre la dirección del campo $E$ y la línea que une las cargas $q_2$ ({q2}nC) y $q_4$ ({q4}nC)?  Exprese su respuesta usando un ángulo en grados, en el rango $0^\\circ\\le\\theta\\le90^\\circ$.

\n

$\\theta=$[[0]]$^\\circ$.

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