Learning Activity #3: PROBLEM SOLVING
Directions: Solve the following problems accurately with complete solutions. (Show the given and solution)
1. A cylindrical metal can has a height of 27 cm and a radius of 11 cm. the electric
field is directed outward along the entire surface of the can (including the top and
bottom), with a uniform magnitude of 4.0 x 105 N/C. How much charge does the
can contain?
2. An insulating sphere with a radius of 20 cm carries a uniform volume charge density of 1.5 x 10-6 C/m3. Find the magnitude of the electric field at a point inside the sphere that lies 8.0 cm from the center.
3. A square metal plate with a thickness of 1.5 cm has no net charge and is placed
in a region of uniform electric field 8.0 x 104 N/C directed perpendicularly to the
plate. Find the resulting surface charge density on each face of the plate.
Problem 1:
Given:
Height of the cylindrical metal can (h) = 27 cm
Radius of the cylindrical metal can (r) = 11 cm
Electric field magnitude (E) = 4.0 x 10^5 N/C
To find:
Charge contained by the can (Q)
Solution:
The electric field is directed outward along the entire surface of the can, including the top and bottom. To find the charge contained by the can, we can use Gauss's law, which states that the electric flux through a closed surface is proportional to the charge enclosed by that surface.
The electric flux (Φ) through the curved surface of the cylindrical can can be calculated using the formula:
Φ = E * A,
where E is the electric field magnitude and A is the area of the curved surface.
The area of the curved surface of a cylinder can be calculated as:
A = 2πrh,
where r is the radius and h is the height.
Substituting the given values:
A = 2 * π * 11 cm * 27 cm
A = 594π cm^2
Now, we can calculate the charge contained by the can using the formula:
Q = Φ / ε0,
where ε0 is the permittivity of free space.
The permittivity of free space is approximately 8.85 x 10^-12 C^2/(N m^2).
Substituting the values:
Q = (E * A) / ε0
Q = (4.0 x 10^5 N/C) * (594π cm^2) / (8.85 x 10^-12 C^2/(N m^2))
Performing the calculations:
Q ≈ 2.0 x 10^-6 C
Therefore, the can contains a charge of approximately 2.0 x 10^-6 C.
Problem 2:
Given:
Radius of the insulating sphere (r) = 20 cm
Volume charge density (ρ) = 1.5 x 10^-6 C/m^3
Distance from the center of the sphere to the point inside (d) = 8.0 cm
To find:
Magnitude of the electric field at the given point inside the sphere (E)
Solution:
To find the magnitude of the electric field at a point inside the sphere, we can use Gauss's law again. Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface.
The electric flux (Φ) through a sphere of radius r can be calculated using the formula:
Φ = E * A,
where E is the electric field magnitude and A is the area of the sphere.
The area of a sphere can be calculated as:
A = 4πr^2.
Substituting the given values:
A = 4 * π * (20 cm)^2
A = 1600π cm^2
The charge enclosed by the sphere can be calculated using the formula:
Q = ρ * V,
where ρ is the volume charge density and V is the volume of the sphere.
The volume of a sphere can be calculated as:
V = (4/3) * π * r^3.
Substituting the given values:
V = (4/3) * π * (20 cm)^3
V = (4/3) * π * 8000 cm^3
The charge enclosed by the sphere is:
Q = (1.5 x 10^-6 C/m^3) * (8000 cm^3)
Q = 0.012 C
Now, we can calculate the electric field using the formula:
E = Q / (4πε0r^2),
where ε0 is the permittivity of free space.
Substituting the values:
E = (0.012 C) / (4π * 8.85 x 10^-12 C^2/(N m^2) * (8 cm)^2)
Performing the calculations:
E ≈ 6.369 x 10^9 N/C
Therefore, the magnitude of the electric field at the point inside the sphere is approximately 6.369 x 10^9 N/C.
Problem 3:
Given:
Thickness of the square metal plate (t) = 1.5 cm
Electric field magnitude (E) = 8.0 x 10^4 N/C
To find:
Surface charge density on each face of the plate (σ)
Solution:
The electric field is directed perpendicularly to the plate, and the plate has no net charge. Therefore, the electric field is balanced by the induced charges on the two faces of the plate.
The surface charge density (σ) can be calculated using the formula:
σ = ε0 * E,
where ε0 is the permApologies for the incomplete response. Here's the complete solution for Problem 3:
Given:
Thickness of the square metal plate (t) = 1.5 cm
Electric field magnitude (E) = 8.0 x 10^4 N/C
To find:
Surface charge density on each face of the plate (σ)
Solution:
The electric field is directed perpendicularly to the plate, and the plate has no net charge. Therefore, the electric field is balanced by the induced charges on the two faces of the plate.
The surface charge density (σ) can be calculated using the formula:
σ = ε0 * E,
where ε0 is the permittivity of free space.
The permittivity of free space is approximately 8.85 x 10^-12 C^2/(N m^2).
Substituting the given values:
σ = (8.85 x 10^-12 C^2/(N m^2)) * (8.0 x 10^4 N/C)
Performing the calculations:
σ ≈ 7.08 x 10^-7 C/m^2
Therefore, the resulting surface charge density on each face of the plate is approximately 7.08 x 10^-7 C/m^2.