Two parallel plates carry uniform charge densities –0.45nC/m2 and +0.45nC/m2 .
Find the magnitude of the electric field between the plates.
Find the magnitude of the acceleration of an electron between these plates.
To find the magnitude of the electric field between the plates, you can use the formula for electric field due to parallel plates. The electric field between two parallel plates carrying uniform charge densities is given by:
E = σ / ε₀
Where:
E is the electric field strength,
σ is the charge density,
ε₀ is the permittivity of free space.
In this case, we have two plates with charge densities of -0.45nC/m² and +0.45nC/m². Note that the charge densities are opposite in sign.
First, we need to convert the charge densities to the proper units. The charge density -0.45nC/m² is equal to -0.45 × 10⁻⁹ C/m², and +0.45nC/m² is equal to +0.45 × 10⁻⁹ C/m².
Next, we need to determine the permittivity of free space, ε₀. The permittivity of free space is approximately 8.85 × 10⁻¹² C²/(N·m²).
Now we can calculate the electric field between the plates.
E = (σ₁ - σ₂) / ε₀
E = (-0.45 × 10⁻⁹ - 0.45 × 10⁻⁹) / (8.85 × 10⁻¹²)
E = -0.9 × 10⁻⁹ / (8.85 × 10⁻¹²)
Calculating this, we find that the magnitude of the electric field between the plates is approximately 101.7 N/C.
To find the magnitude of the acceleration of an electron between these plates, we can use the equation:
a = eE / m
Where:
a is the acceleration,
e is the charge of an electron (approximately -1.6 × 10⁻¹⁹ C),
E is the electric field strength, and
m is the mass of an electron (approximately 9.1 × 10⁻³¹ kg).
Plugging in the values:
a = (-1.6 × 10⁻¹⁹ C) * (101.7 N/C) / (9.1 × 10⁻³¹ kg)
Calculating this, we find that the magnitude of the acceleration of an electron between the plates is approximately 1.79 × 10¹⁴ m/s².