Respuesta :
Answer:
(a) Electric field at 0.250 m is zero.
(b) Electric field at 2.90 m is zero.
(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.
(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.
Explanation:
Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.
Electric field inside and outside the metal sphere is :
E = 0 for r ≤ R ( inside )
= [tex]\frac{KQ}{r^{2} }[/tex] for r > R ( outside )
Here K is electric constant and r is the distance from the center of the metal sphere.
(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.
(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.
(c) Electric field at 3.10 m is given by the relation as r > R :
E = [tex]\frac{KQ}{r^{2} }[/tex]
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.
E = - [tex]\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }[/tex]
E = - 5.15 x 10³ V/m
(d) Electric field at 8.00 m is given by the relation as r > R :
E = [tex]\frac{KQ}{r^{2} }[/tex]
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.
E = - [tex]\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }[/tex]
E = - 0.77 x 10³ V/m
The characteristics of the electric field and Gauss's law allow finding the results for the electric field at various distances from a charged metallic sphere are:
a) Distance r = 0.250 m, E = 0
b) Distance r = 2.90 m, E = 0
c) Distance r = 3.10 m, E = -5.15 10³ V / m
d) Distance r = 8.00 m, E = -0.77 10³ V / m
The electric field is defined as the relationship between the electric force on a test charge.
The materials can be classified according to the mobility of their electrical charges into two types;
- Insulators. They are materials where the formed charge cannot move.
- Conductives. They are the materials where the charges formed can move freely, due to the electrical repulsion between charges of the same sign, they are moved to the surface of the material.
They indicate that we have a metal sphere, therefore the charges are free to move, due to the electrical repulsion, it moves towards the surface.
Using Gauss's law that states that the electric field flux is proportional to the charge inside the surface. We conclude that in the sphere there are no charges inside, therefore for smaller radii and the radius of the sphere the electric field is zero.
For radii greater than the radius of the sphere, according to Gauss's law, it is compiled as if all the charge were at its center, therefore the expression for the electuary field is:
[tex]E = k \frac{q}{r^2 }[/tex]
Where k is the Coulomb constant, q the charge on the sphere, and r the distance from the center of the sphere to the point of interest.
Case of r = 3.10 m.
Let's calculate
[tex]E = 9 \ 10^9 \frac{(-5.50)}{3.10^2}[/tex]
E = -5.15 10³ V / m
Case of r = 8.0 m.
[tex]e = 9 \ 10^9 \frac{(-5.50)}{8.0^2 }[/tex]
E = -0.77 10³ V / m
The negative sign indicates that the field is directed radially towards the center of the sphere.
In conclusion, using the characteristics of the electric field and Gauss's law we can find the results for the electric field at various distances from a charged metallic sphere are:
a) Distance r = 0.250 m, E = 0
b) Distance r = 2.90 m, E = 0
c) Distance r = 3.10 m, E = -5.15 10³ V / m
d) Distance r = 8.00 m, E = -0.77 10³ V / m
Learn more about the electric field here: brainly.com/question/14372859
