Two plates, each of area 3.40*10-4 m2, are used to construct a parallel-plate capacitor with capacitance 1.20 pF.

(a) Find the necessary separation distance.
d = ?m

(b) If the positive plate is to hold a charge of 4.70* 10-12 C, find the charge density.
σ = ?C/m2

(c) Find the electric field between the plates.
E = ?N/C

(d) What voltage battery should be attached to the plate to obtain the preceding results?
V = ?V

To find the necessary separation distance between the plates, we can use the formula for capacitance of a parallel-plate capacitor:

C = ε₀ * (A / d)

Where:
C = capacitance (1.20 pF = 1.20 × 10^(-12) F)
ε₀ = permittivity of free space (8.85 × 10^(-12) F/m)
A = area of each plate (3.40 × 10^(-4) m²)
d = separation distance (unknown)

(a) Rearrange the formula to solve for d:

d = ε₀ * (A / C)

Substituting the known values:

d = (8.85 × 10^(-12) F/m) * (3.40 × 10^(-4) m²) / (1.20 × 10^(-12) F)

Calculating the result:

d = 8.85 × 3.40 / 1.20

d ≈ 24.99 × 10^(-5) m
d ≈ 2.499 × 10^(-3) m
d ≈ 2.499 mm

So, the necessary separation distance is approximately 2.499 mm.

(b) To find the charge density (σ) on the positive plate, we simply divide the charge on the plate by its area:

σ = Q / A

Where:
σ = charge density (unknown)
Q = charge on the positive plate (+4.70 × 10^(-12) C)
A = area of the plate (3.40 × 10^(-4) m²)

Substituting the known values:

σ = (4.70 × 10^(-12) C) / (3.40 × 10^(-4) m²)

Calculating the result:

σ = 4.70 / 3.40 × 10^(-12-4) C/m²

σ ≈ 1.382 × 10^(-4) C/m²

So, the charge density is approximately 1.382 × 10^(-4) C/m².

(c) The electric field between the plates of a parallel-plate capacitor is given by the formula:

E = σ / ε₀

Where:
E = electric field strength (unknown)
σ = charge density (1.382 × 10^(-4) C/m²)
ε₀ = permittivity of free space (8.85 × 10^(-12) F/m)

Substituting the known values:

E = (1.382 × 10^(-4) C/m²) / (8.85 × 10^(-12) F/m)

Calculating the result:

E = 1.382 / 8.85 × 10^(-4-12) C/F

E ≈ 0.156 × 10^(-8) N/C
E ≈ 1.56 × 10^(-9) N/C

So, the electric field strength between the plates is approximately 1.56 × 10^(-9) N/C.

(d) To find the voltage required to obtain the previous results, we can use the formula for electric field strength:

E = V / d

Where:
E = electric field strength (1.56 × 10^(-9) N/C)
V = voltage (unknown)
d = separation distance (2.499 × 10^(-3) m)

Rearranging the formula to solve for V:

V = E * d

Substituting the known values:

V = (1.56 × 10^(-9) N/C) * (2.499 × 10^(-3) m)

Calculating the result:

V = 1.56 × 2.499 × 10^(-9-3) Nm/C
V ≈ 0.00389 Nm/C
V ≈ 3.89 × 10^(-3) Nm/C
V ≈ 3.89 mV

So, the voltage required is approximately 3.89 mV.

(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.

In the figure here, a red car and a green car move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 223 m. If the red car has a constant velocity of 22.0 km/h, the cars pass each other at x = 44.7 m. On the other hand, if the red car has a constant velocity of 44.0 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity (in km/h) and (b) the (constant) acceleration (in m/s2) of the green car?

For (a), find the formula for C in terms of epsilon, Area and plate separation. For (b), divide the charge by the area. For (c), use Gauss law and the charge density for the E field . For (d), use V = E*d

Show your work if you need further assistance.