Answer:
r = 2.55 m
Explanation:
Given that,
First charge, q₁ = 32.2 μC
Second charge, q₂ = + 12.3 μC
The force between charges, F = 0.544 N
We need to find the distance between charges. The force between two charges is given by the formula as follows :
So, the charges are 2.55 m apart.
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