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Answer:
x' = 1.01 m
Explanation:
given,
mass suspended on the spring, m = 0.40 Kg
stretches to distance, x = 10 cm = 0. 1 m
now,
we know
m g = k x
where k is spring constant
0.4 x 9.8 = k x 0.1
k = 39.2 N/m
now, when second mass is attached to the spring work is equal to 20 J
work done by the spring is equal to
x'² = 1.0204
x' = 1.01 m
hence, the spring is stretched to 1.01 m from the second mass.
Answer:
1) The total charge of the top plate is 0.008 C
b) The total charge of the bottom plate is -0.008 C
2) The electric field at the point exactly midway between the plates is 0
3) The electric field between plates is approximately 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N
Explanation:
The given parameters of the parallel plate capacitor are;
The dimensions of the plates = 4 × 2 cm
The distance between the plates = 10 cm
The surface charge density of the top plate, σ₁ = 10 C/m²
The surface charge density of the bottom plate, σ₂ = -10 C/m²
The surface area, A = 0.04 m × 0.02 m = 0.0008 m²
1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C
b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C
2) The electrical field at the point exactly midway between the plates is given as follows;
Therefore, we have;
The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m
The electric field at the point exactly midway between the plates, = 0
3) The electric field, 'E', between plates is given as follows;
E ≈ 1.1294 × 10¹² N/C
The electric field between plates, E ≈ 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates
The charge on an electron, e = -1.6 × 10⁻¹⁹ C
The force on an electron in the middle of the two plates, = E × e
∴ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N
The force on an electron in the middle of the two plates, ≈ 1.807 × 10⁻⁷ N