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Ganezh [65]
3 years ago
12

When applying Gauss' law for a capacitor containing a dielectric, which of the following statements is false? (Latter part of se

ction 6.)
A) The electric field vector is multiplied by κ, the dielectric constant.
B) The charge enclosed by the Gaussian surface includes both the free charge and the induced surface charges.
C) Gauss' law for a dielectric includes the ability to account for varying dielectric constant over the Gaussian surface.
D) Gauss' law for a dielectric is the most general form of Gauss' law.
E) Gauss' law for a dielectric can be applied to many types of capacitors.
Physics
1 answer:
Irina-Kira [14]3 years ago
3 0

Answer:a

Explanation:

Introducing a dielectric inside a capacitor reduces the Electric field . Charge develops on dielectric such that it is opposite to the corresponding face . For example Positive charge of capacitor induces negative charge on dielectric which will create an Electric Field in opposite direction.

So Electric Field decreased by a factor of \frac{1}{K} where

K=dielectric\ magnitude    

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Cual es la fuerza electrica sobre el electrón (-1.6 x 10¹⁹c) de un atomo de hidrógeno ejercida por el protón (1.6 x 10¹⁹c)? Supó
kkurt [141]

Answer:

La  fuerza eléctrica es -8.2*10⁻⁸ N

Explanation:

El enunciado correcto es: <em>¿Cuál es la fuerza eléctrica sobre el electrón (-1.6 x 10⁻¹⁹c) de un átomo de hidrógeno ejercida por el protón (1.6 x 10⁻¹⁹c)? Supóngase que la distancia entre el electrón y el protón es de 5.3 x 10⁻¹¹ m</em>

Entre dos o más cargas aparece una fuerza denominada fuerza eléctrica. Su valor depende del valor de las cargas y de la distancia que las separa, mientras que su signo depende del signo de cada carga. Las cargas del mismo signo se repelen entre sí, mientras que las de distinto signo se atraen.

La fuerza eléctrica con la que se atraen o repelen dos cargas puntuales en reposo es directamente proporcional al producto de las mismas e inversamente proporcional al cuadrado de la distancia que las separa:

F=K*\frac{q1*q2}{d^{2} }

donde:

  • F es la fuerza eléctrica de atracción o repulsión. En el Sistema Internacional (S.I.) se mide en Newtons (N).
  • q1 y q2 son lo valores de las dos cargas puntuales. En el S.I. se miden en Culombios (C).
  • d es el valor de la distancia que las separa. En el S.I. se mide en metros (m).
  • K es una constante de proporcionalidad llamada constante de la ley de Coulomb. Depende del medio en el que se encuentren las cargas. Para el vacío K tiene un valor aproximadamente de 9*10⁹ \frac{N*m^{2} }{C^{2} }.

En este caso:

  • F=?
  • K= 9*10⁹ \frac{N*m^{2} }{C^{2} }
  • q1= -1.6*10⁻¹⁹ C
  • q2= 1.6*10⁻¹⁹ C
  • d= 5.3*10⁻¹¹ m

Reemplazando:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{(-1.6*10^{19} C)*(1.6*10^{19} C)}{(5.3*10^{-11} )^{2} }

Resolviendo:

F= -8.2*10⁻⁸ N

<u><em>La  fuerza eléctrica es -8.2*10⁻⁸ N</em></u>

6 0
4 years ago
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