Problema para la siguiente configuracion de cargas situadas...

Problema: Para la siguiente configuración de cargas situadas en los vértices de un cubo de lados $s$. Calcula el campo eléctrico en el centro de una de sus caras. a) (25Puntos) Dibuja un diagrama vectorial totalmente detallado de las componentes de los vectores a considerar en el punto de interés (centro de una de las caras). Puedes ayudarte de una herramienta digital b) (25Puntos) Escribe la expresión para $\vec{E}$ en términos de las constantes: $q, s, k$ c) (25Puntos) Si $q = 1\mu C, s = 1cm$ dá un valor numérico para este campo eléctrico. d) (25Puntos) ¿Cuál es la dirección del campo eléctrico en el centro de la cara superior del cubo? Punto A

Transcript

00:01 In this problem we have to find the potential energy of the system.

00:06 So we have 4 charges, plus q, minus q, plus q and minus q.

00:18 The diagonal is of length, r each, and diagonal is equal to square root 2 times the side length.

00:33 Now potential energy of the system is due to the pair of charges.

00:41 So for four charges we will have six pairs so potential energy of system total potential energy due to charge at a and b which will be columns constant times the product of charges q times minus q divided by separation between them which is equal to a then we have potential due to charges at b and c so this will be k times product of charges then we have potential due to charge at c and d this is also k times product of charges divided by separation between them then we have potential due to charges at c at a and d which is column's constant times product of charges divided by separation between them then we have potential energy due to charges at a and c so this is k times q times q product of charges divided by separation between them which is equal to the length of diagonal r and r is equal to square root 2a similarly we have potential energy due to charger at b and d so columns constant times the product of charges divided by separation between them square root 2a so we have 4 k q square minus divided by a plus 2 k q square square square square root 2 a k q square upon a is common so we have minus 4 plus 2 upon square root 2 k is columns constant 9 times 10 to the power 9 q with charge which is 2 micropoleums 2 times 10 to the power minus 6 columns a is equal to 5 cm which is 0 .05 meters times minus 4 2 upon square root 2 is square root 2 so we get this is square minus 1 .86 joule this is the required potential energy of the system now for part b we need to find the potential at the center so the center of the square is at distance d from each corner, a, b, c and d, center is at o.

04:03 So the potential at center is total potential, which is potential due to charge at a, plus potential due to charge at b, plus potential due to charge at c, plus potential due to charge at d, potential due to charge at a at the center...

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