A hollow metal sphere with a radius of 5 cm is uniformly charged with a total charge of +5 µC. The sphere is electrically isolated and located in a vacuum.
(a) Determine the electric field both inside and outside the sphere.
(b) If an electron is released from rest at a point 10 cm away from the center of the sphere, in what direction and with what acceleration will it move?
(c) Calculate the electric potential at a point 20 cm away from the center of the sphere.
(a) To determine the electric field both inside and outside the sphere, we can make use of Gauss's Law. Gauss's Law states that the total electric flux through a closed surface is proportional to the total charge enclosed within that surface.
For a spherical surface with radius r > R (outside the sphere), the electric field on the surface will be radial, but its magnitude will be constant:
E = k * (Qenclosed / r²)
where E is the electric field, k is Coulomb's constant (8.99 x 10^9 N*m²/C²), and Qenclosed is the total charge enclosed.
The total charge enclosed for points outside the sphere is equal to the total charge of the sphere, so we have:
E(r > R) = k * (Q / r²)
Substituting the given values, we have:
E(r > R) = (8.99 x 10^9 N*m²/C²) * (5 x 10^-6 C) / (0.1 m)²
E(r > R) = 4.495 x 10^7 N/C
For a spherical surface with radius r < R (inside the sphere), the electric field on the surface will also be radial, but its magnitude will be zero. This is because the total charge enclosed within the sphere is zero, due to its hollow configuration.
Therefore, the electric field inside the sphere is zero:
E(r < R) = 0 N/C
(b) To determine the direction and acceleration of an electron released from rest at a point 10 cm away from the center of the sphere, we need to consider the force acting on the electron. The force experienced by a charged particle in an electric field is given by:
F = q * E
where F is the force, q is the charge of the particle, and E is the electric field.
In this case, the charge of the electron is -1.6 x 10^-19 C. At a distance of 10 cm from the center of the sphere, the electric field is:
E = (8.99 x 10^9 N*m²/C²) * (5 x 10^-6 C) / (0.1 m)²
E = 4.495 x 10^7 N/C
Therefore, the force acting on the electron is:
F = (-1.6 x 10^-19 C) * (4.495 x 10^7 N/C)
F = -7.192 x 10^-12 N
The negative sign indicates that the force is attractive towards the positively charged sphere.
Using Newton's second law, we know that the force acting on an object is equal to its mass multiplied by its acceleration:
F = m * a
Since the electron's mass is 9.1 x 10^-31 kg, we can solve for the acceleration:
a = F / m
a = (-7.192 x 10^-12 N) / (9.1 x 10^-31 kg)
a = -7.90 x 10^18 m/s²
The negative sign indicates that the acceleration is directed towards the positively charged sphere.
(c) To calculate the electric potential at a point 20 cm away from the center of the sphere, we can use the equation for electric potential due to a point charge:
V = k * (Q / r)
where V is the electric potential, k is Coulomb's constant, Q is the total charge, and r is the distance from the center of the sphere.
Plugging in the given values, we have:
V = (8.99 x 10^9 N*m²/C²) * (5 x 10^-6 C) / (0.2 m)
V = 2.2475 x 10^6 V
Therefore, the electric potential at a point 20 cm away from the center of the sphere is 2.2475 x 10^6 V.
Note: In this solution, we assumed that the electric field is radially symmetric around the sphere due to its symmetry.