An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. If a 19.2-V potential difference is applied to these plates, calculate the following:

(a) the electric field between the plates, magnitude (in kV/m) and direction
(b) the capacitance (pF)
(c) the charge on each plate (pC)

(a) E = V/d = 19.2 V/.0017m

= 112,900 V/m = 112.9 kV/m

(b) C = (epsilon0)*A/d
= 8.85*10^-12 Farad/m*7.6*10^-4
m^2/0.0017 m
= 3.96*10^-11 F
= 39.6 pF
(c) Q = C*V = 39.6*10^-12F*19.2 V
= 6.41*10^-10 C = _____ pC

(a) Oh boy, let's charge up for some electric field calculations! To find the magnitude of the electric field (E) between the plates, we can use the formula E = V/d, where V is the potential difference and d is the distance between the plates. In this case, V = 19.2 V and d = 1.70 mm = 0.00170 m. Plugging these values into the formula, we get E = 19.2 V / 0.00170 m = 11324.7 V/m. Now, let's convert that to kV/m by dividing by 1000. So, the magnitude of the electric field between the plates is 11.32 kV/m. As for the direction, the electric field always points from the positive plate to the negative plate.

(b) Now, let's get our "cap" together and calculate the capacitance (C). The capacitance of a parallel-plate capacitor can be calculated using the formula C = ε₀ * (A/d), where ε₀ is the permittivity of free space, A is the area of one plate, and d is the distance between the plates. The permittivity of free space, ε₀, is approximately 8.85 × 10^-12 F/m. Given that A = 7.60 cm² = 7.60 × 10^-4 m² and d = 1.70 mm = 0.00170 m, we can plug these values into the formula to calculate the capacitance. So, C = 8.85 × 10^-12 F/m * (7.60 × 10^-4 m² / 0.00170 m) = 40.768 F. To convert this to picofarads (pF), we multiply by 10^12. Therefore, the capacitance of the air-filled capacitor is approximately 40.768 pF.

(c) Now, let's find out how much charge is partying on each plate. To calculate the charge (Q) on each plate, we can use the formula Q = C * V, where C is the capacitance and V is the potential difference. In this case, C = 40.768 pF = 40.768 × 10^-12 F and V = 19.2 V. Plugging these values into the formula, we get Q = 40.768 × 10^-12 F * 19.2 V ≈ 7.841 pC. Therefore, the charge on each plate of the air-filled capacitor is approximately 7.841 pC.

To solve these problems, we can use the equations related to capacitors.

(a) The electric field between the plates can be calculated using the equation:

Electric field (E) = Voltage (V) / Distance between plates (d)

Given:
Voltage (V) = 19.2 V
Distance between plates (d) = 1.70 mm = 1.70 x 10^-3 m

Substituting these values into the equation, we get:

E = 19.2 V / (1.70 x 10^-3 m)

Calculating this gives:

E = 1.13 x 10^4 V/m

The direction of the electric field is from the positive plate to the negative plate.

(b) The capacitance (C) can be calculated using the equation:

Capacitance (C) = (Permittivity of free space) x (Area of each plate) / Distance between plates

Given:
Area of each plate (A) = 7.60 cm^2 = 7.60 x 10^-4 m^2
Distance between plates (d) = 1.70 mm = 1.70 x 10^-3 m
Permittivity of free space (ε0) = 8.85 x 10^-12 F/m (constant)

Substituting these values into the equation, we get:

C = (8.85 x 10^-12 F/m) x (7.60 x 10^-4 m^2) / (1.70 x 10^-3 m)

Calculating this gives:

C = 3.95 x 10^-12 F = 3.95 pF

So, the capacitance is 3.95 pF.

(c) The charge on each plate can be calculated using the equation:

Charge (Q) = Capacitance (C) x Voltage (V)

Given:
Capacitance (C) = 3.95 x 10^-12 F
Voltage (V) = 19.2 V

Substituting these values into the equation, we get:

Q = (3.95 x 10^-12 F) x (19.2 V)

Calculating this gives:

Q = 7.59 x 10^-11 C = 75.9 pC

So, the charge on each plate is 75.9 pC.

To calculate the answers to the given questions, you need to use the formulas related to capacitor properties. Here's how you can approach each part:

(a) The electric field between the plates, magnitude, and direction:
The electric field between the plates of a parallel plate capacitor can be calculated using the formula:

Electric Field (E) = Voltage (V) / Distance between plates (d)

Given:
Voltage (V) = 19.2 V
Distance between plates (d) = 1.70 mm = 1.70 × 10^(-3) m

Plugging in the values, we get:
Electric Field (E) = 19.2 V / (1.70 × 10^(-3) m)

Now, simply calculate the value of the electric field. The magnitude of the electric field will be positive, indicating its direction is from the positive plate towards the negative plate.

(b) The capacitance:
The capacitance of a parallel plate capacitor can be calculated using the formula:

Capacitance (C) = (Permittivity of free space (ε₀) × Area of the plates (A)) / Distance between plates (d)

Given:
Area of the plates (A) = 7.60 cm² = 7.60 × 10^(-4) m²
Permittivity of free space (ε₀) = 8.85 × 10^(-12) F/m
Distance between plates (d) = 1.70 mm = 1.70 × 10^(-3) m

Plugging in the values, we get:
Capacitance (C) = (8.85 × 10^(-12) F/m × 7.60 × 10^(-4) m²) / (1.70 × 10^(-3) m)

Now, simply calculate the value of the capacitance.

(c) The charge on each plate:
The charge on each plate of a capacitor can be calculated using the formula:

Charge (Q) = Capacitance (C) × Voltage (V)

Given:
Capacitance (C) - calculated in part (b) before
Voltage (V) = 19.2 V

Plugging in the values, we get:
Charge (Q) = Capacitance (C) × Voltage (V)

Now, simply calculate the value of the charge on each plate.

Please let me know if you need further assistance with the calculations.