Answer:
Explanation:
a )
in the regions r < R₁
charge q inside sphere of radius R₁ = 0
Applying gauss's law for electric field E at distance r <R₁
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q / ε₀ = 0 / ε₀
E = 0
Applying gauss's law for electric field E at distance R₁ < r < R₂ .
charge q inside sphere of radius R₁ = q₁
Applying gauss's law for electric field E at distance R₁ < r < R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q₁ / ε₀
E = q₁ / 4πε₀
in the regions r> R₂
charge q inside sphere of radius R₂ = (q₁ + q₂)
Applying gauss's law for electric field E at distance r > R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = (q₁ + q₂) / ε₀
E = (q₁ + q₂) /4π ε₀
b )
For electric flux to be zero at r > R₂
(q₁ + q₂) /4π ε₀ = 0
q₁ + q₂ = 0
q₁ / q₂ = - 1 .
Answer:
La magnitud de la fuerza de atracción entre las cargas es 0.648 newtons.
Explanation:
La fuerza eléctrica de atracción entre dos partículas cargadas eléctricamente es determinada por la ley de Coulomb, la cual dice que la fuerza eléctrica es directamente proporcional a la carga eléctrica e inversamente proporcional de la distancia entre las partículas. Matemáticamente hablando, la formula es descrita a continuación:
(1)
Donde:
- Fuerza electrostática, medida en newtons.
- Constante electrostática, medida en newton-metros cuadrados por Coulomb cuadrado.
, - Cargas eléctricas, medidas en Coulomb.
- Distancia entre partículas, medidas en metros.
Si sabemos que , , y , entonces la magnitud de la fuerza de atracción entre las cargas es:
La magnitud de la fuerza de atracción entre las cargas es 0.648 newtons.
Friction is what's letting gymnast control his weight by having a tight grip on the bar. Without friction it would be like them trying hold on to ice.
Because any other radiation can be harmful<span />
Answer:
The particle comes to rest before reaching the position x=4m.
Explanation:
When the object is at x=0m, all of its energy is kinetic energy. Using the equation for kinetic energy yields KE=1/2mv^2=(12)(2)(3)^2=9J. Using the given equation for potential energy when the object is at x=4m yields U=4x^2=4(4)^2=64J. Since the system only has 9J of energy, the object comes to rest before reaching x=4m.