(a) To find the necessary separation distance (d), we can use the formula for the capacitance of a parallel-plate capacitor:
C = (ε₀ * A) / d,
where C is the capacitance, ε₀ is the permittivity of free space (8.85 x 10⁻¹² F/m), A is the area of each plate, and d is the separation distance.
Rearranging the formula, we get:
d = (ε₀ * A) / C.
Substituting the given values, we have:
d = (8.85 x 10⁻¹² F/m * 3.40 x 10⁻⁴ m²) / 1.20 x 10⁻¹² F.
Calculating this expression, we find:
d = 1.060 m.
Therefore, the necessary separation distance is 1.060 m.
(b) The charge density (σ) can be obtained by dividing the charge (Q) by the area (A):
σ = Q / A.
Substituting the given values, we get:
σ = (4.70 x 10⁻¹² C) / (3.40 x 10⁻⁴ m²).
Calculating this expression, we find:
σ ≈ 1.38 x 10⁻⁵ C/m².
Therefore, the charge density is approximately 1.38 x 10⁻⁵ C/m².
(c) The electric field (E) between the plates can be calculated using the equation:
E = V / d,
where V is the voltage and d is the separation distance.
In this case, we haven't been given the exact voltage, so we'll skip this step for now.
(d) To calculate the required voltage (V) for the electric field, we can rearrange the previous equation:
V = E * d.
Substituting the given values, we get:
V = (E * d).
Now, we need the value of the electric field (E) to complete the calculation, but it hasn't been provided.
To find the electric field, we can use the formula:
E = σ / (ε₀).
Substituting the given values, we get:
E = (1.38 x 10⁻⁵ C/m²) / (8.85 x 10⁻¹² F/m).
Calculating this expression, we find:
E ≈ 1.56 x 10⁷ N/C.
Now, we can substitute this value for E into the previous equation to find V:
V = (1.56 x 10⁷ N/C) * (1.060 m).
Calculating this expression, we find:
V ≈ 1.66 x 10⁷ V.
Therefore, a voltage of approximately 1.66 x 10⁷ V should be attached to the plate to obtain the desired results.