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WARRIOR [948]
3 years ago
11

Four charges are on the four corners of a square. Q1 = +5μC, Q2 = -10μC, Q3 = +5μC, Q4 = -10μC. The side length of the square is

3 m. What is the magnitude of the net electric field at the center of the square?
Please show the equation(s) used and your work.
Physics
1 answer:
Marat540 [252]3 years ago
3 0

Answer:

Explanation:

Electric field due to a point charge Q at a point at distance d is given by the relation

E = \frac{K\times Q}{d^2}

Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.

The charges are situated on the corners of a square in such a way that

equal charges of Q1 and Q3 are situated on the diametrically  opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two  charges will be zero.  

On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero

Overall, net field due to all the four charges will be zero

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Answer:

  • \bf\pink{10\:m/s}

Explanation:

<h2><u>Given</u> :-</h2>

  • \sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}
  • \sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}

<h2><u>To Find</u> :-</h2>

  • \sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}

<h2><u>Formula to be used</u> :-</h2>

  • \sf\blue{K.E. = \dfrac{1}{2} mv^{2}}

Where,

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<h2><u>Solution</u> :-</h2>

\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}

\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}

\to\:\:\sf\green{5000 = 50 \times v^{2}}

\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}

\to\:\:\sf\purple{100 = v^{2}}

\to\:\:\sf\red{\sqrt{100} = v}

\to\:\:\sf\orange{ 10 = v}

\to\:\:\bf\pink{v = 10\:m/s}

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La tension que se transmite en la cuerda BD es de 75 lb. Calcula el momento de fuerza generada por la cuerda respecto al punto C
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Answer:

Mc = 1920[lb*in]

Explanation:

Para poder solucionar este problema debemos realizar un análisis estático, por tal motivo lo primero es realizar un diagrama de cuerpo libre con las respectivas fuerzas actuando sobre la barra ABC. DE igual manera calcular la geometría de la configuración mostrada.

El diagrama de cuerpo libre se puede ver en la imagen adjunta, con la solución de este problema.

Lo primero es determinar el angulo t, el cual por medio de las propiedades del triangulo rectángulo se puede determinar.

Con este angulo (t) ya determinado, fijamos la atención en el triangulo BCD, este triangulo no es rectángulo, pero por medio de la ley de senos podemos determinar el angulo omega.

Después de determinar el angulo omega, restamos el angulo (t) para poder determinar el angulo (a).

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El momento en el punto C es de 1920 [Lb*in].

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