Two circular metal disc each of 3.00 cm diamter are 1.5 mm apart make a capacitor.

A)What is the capacitance of this capacitor if it is in series to a 25 V battery?
B)What is the magnitude charge developing on each plate?
C)While the batter remains connected, the plates are pulled apart until the 2.5mm? Does the capacitance of capacitor change?is so what is it now?
D)What potential difference develops between the plates?
E)What charge develops on each plate?
F)Battery is disconnected and then plates are pushed together to 1.5mm.(think about whether it will be the charge on the plates or the potential across the plates that will remain constant in this scenario, as opposed to the situation above)
F) What is the capacitance now?
G)What potential difference develops between the plates?
J) how much energy is stored in the capacitor?
K) if at this time the capacitor is allowed to dischared through a 15 ohm resistor what will be the time contact fo this RC circuit?
L) how long will it be before half is discharged?

A) The capacitance of a capacitor is given by the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. In this case, the diameter of each circular plate is 3.00 cm, so the radius is 1.50 cm or 0.015 m. The area of each plate is A = πr² = π(0.015)² = 0.00070685 m². The distance between the plates is 1.5 mm or 0.0015 m. Therefore, the capacitance of the capacitor is C = (8.85 x 10^(-12) F/m)(0.00070685 m²)/(0.0015 m) = 3.30 x 10^(-10) F.

B) When the capacitor is connected to a 25 V battery in series, the magnitude of the charge on each plate can be calculated using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor. In this case, the capacitance is 3.30 x 10^(-10) F and the potential difference is 25V. Therefore, the magnitude of the charge on each plate is Q = (3.30 x 10^(-10) F)(25V) = 8.25 x 10^(-9) C.

C) When the plates are pulled apart until the separation distance becomes 2.5 mm or 0.0025 m, the capacitance of the capacitor changes. The new capacitance can be calculated using the same formula as before: C' = (8.85 x 10^(-12) F/m)(0.00070685 m²)/(0.0025 m) = 2.484 x 10^(-10) F.

D) The potential difference that develops between the plates can be calculated using the equation V = Q/C, where V is the potential difference, Q is the charge, and C is the capacitance. In this case, the charge on each plate is 8.25 x 10^(-9) C, and the capacitance is 3.30 x 10^(-10) F. Therefore, the potential difference is V = (8.25 x 10^(-9) C) / (3.30 x 10^(-10) F) = 25V.

E) The charge on each plate remains unchanged at 8.25 x 10^(-9) C.

F) When the battery is disconnected and the plates are pushed back together to a separation distance of 1.5 mm or 0.0015 m, the capacitance of the capacitor remains the same as in the initial configuration, which is 3.30 x 10^(-10) F.

F) The capacitance remains 3.30 x 10^(-10) F.

G) When the plates are pushed together to a separation distance of 1.5 mm or 0.0015 m, the potential difference across the plates remains the same at 25V.

J) The energy stored in a capacitor can be calculated using the equation U = 1/2CV², where U is the energy stored, C is the capacitance, and V is the potential difference across the capacitor. In this case, the capacitance is 3.30 x 10^(-10) F and the potential difference is 25V. Therefore, the energy stored in the capacitor is U = 1/2(3.30 x 10^(-10) F)(25V)² = 8.125 x 10^(-8) J.

K) The time constant (τ) of an RC circuit can be calculated using the equation τ = RC, where R is the resistance and C is the capacitance. In this case, the resistance is 15 ohms and the capacitance is 3.30 x 10^(-10) F. Therefore, the time constant is τ = (15 ohms)(3.30 x 10^(-10) F) = 4.95 x 10^(-9) s.

L) The time it takes for half of the charge to be discharged in an RC circuit is approximately equal to 0.693 times the time constant. In this case, the time constant is 4.95 x 10^(-9) s. Therefore, the time it takes for half of the charge to be discharged is (0.693)(4.95 x 10^(-9) s) = 3.42 x 10^(-9) s.

To solve the given questions, we'll need to use the formulas and equations related to capacitors. Let's go through each question step by step:

A) To find the capacitance of the capacitor, we can use the formula 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 one plate, and d is the separation between the plates.

Since each disc has a diameter of 3.00 cm, the radius (r) would be 1.50 cm or 0.015 m. Therefore, the area of each plate is A = π * r^2 = π * (0.015 m)^2 = 0.0007068 m^2.

The separation between the plates is given as 1.5 mm or 0.0015 m.

Using the formula C = ε0 * A / d, we can calculate the capacitance:
C = (8.85 x 10^-12 F/m) * (0.0007068 m^2) / 0.0015 m = 4.142 x 10^-9 F

Thus, the capacitance of this capacitor is 4.142 nanofarads (nF).

B) The magnitude of charge on each plate can be found using the equation Q = C * V, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.

Given that the capacitor is in series with a 25 V battery and the capacitance is 4.142 nF, we can calculate the charge:
Q = (4.142 x 10^-9 F) * 25 V = 1.0355 x 10^-7 C

So, the magnitude charge developing on each plate is approximately 1.0355 x 10^-7 C.

C) When the plates are pulled apart until the separation becomes 2.5 mm or 0.0025 m, the capacitance of the capacitor changes. The new capacitance can be calculated using the same formula: C = ε0 * A / d.

Now, the separation (d) is 0.0025 m. Plugging in the values, we get:
C' = (8.85 x 10^-12 F/m) * (0.0007068 m^2) / 0.0025 m ≈ 2.472 x 10^-9 F

The new capacitance of the capacitor is approximately 2.472 nF.

D) The potential difference (V) between the plates of the capacitor can be found using the formula V = Q / C. Recall that the charge (Q) remains constant.

So, the potential difference is:
V = (1.0355 x 10^-7 C) / (4.142 x 10^-9 F) ≈ 24.98 V

The potential difference between the plates is approximately 24.98 V.

E) The charge on each plate remains the same, as the charge is conserved. Therefore, the charge on each plate is still approximately 1.0355 x 10^-7 C.

F) In this scenario, where the battery is disconnected and the plates are pushed together to 1.5 mm or 0.0015 m, the charge on the plates remains constant. Therefore, it is the potential across the plates that will remain unchanged.

F) The capacitance of the capacitor can be found using the same formula: C = ε0 * A / d.

Now, the separation (d) is 0.0015 m. Plugging in the values, we get:
C'' = (8.85 x 10^-12 F/m) * (0.0007068 m^2) / 0.0015 m ≈ 4.142 x 10^-9 F

The capacitance of the capacitor remains the same, approximately 4.142 nF.

G) The potential difference across the plates can be found using the formula V = Q / C.

Since the charge (Q) remains constant and the capacitance (C) is the same as before, the potential difference also remains the same. Therefore, the potential difference is approximately 24.98 V.

J) The energy stored in the capacitor can be calculated using the formula U = (1/2) * C * V^2, where U is the energy stored, C is the capacitance, and V is the voltage across the capacitor.

Plugging in the values, we get:
U = (1/2) * (4.142 x 10^-9 F) * (24.98 V)^2 ≈ 1.299 x 10^-5 J

The energy stored in the capacitor is approximately 1.299 x 10^-5 Joules.

K) To find the time constant (τ) for the RC circuit, we need to use the formula τ = R * C, where R is the resistance and C is the capacitance.

Given that the resistance is 15 ohms and the capacitance is 4.142 nF, we can calculate the time constant:
τ = (15 Ω) * (4.142 x 10^-9 F) ≈ 6.213 x 10^-8 s

The time constant for the RC circuit is approximately 6.213 x 10^-8 seconds.

L) Half of the charge will be discharged when the voltage across the capacitor reaches approximately 0.5 times the initial voltage (25 V). Since the potential difference across the plates decreases exponentially with time, we can use the equation V(t) = V0 * e^(-t/τ), where V0 is the initial voltage, t is the time, and τ is the time constant.

Setting V(t) to 0.5 times V0:
0.5 * 25 V = 25 V * e^(-t/6.213 x 10^-8 s)

Simplifying and solving for t:
e^(-t/6.213 x 10^-8 s) = 0.5
(-t/6.213 x 10^-8 s) = ln(0.5)
t ≈ -6.213 x 10^-8 s * ln(0.5)
t ≈ 4.328 x 10^-8 s

Therefore, it will take approximately 4.328 x 10^-8 seconds for half of the charge to be discharged.