Two 2.5-cm-diameter electrodes with a 0.10-mm-thick sheet of Teflon between them are attached to a 9.0V battery. Without disconnecting the battery, the Teflon is removed.
a) What is the charge before the Teflon is removed? Q= C
b) What is the potential difference before the Teflon is removed? V= V
c) What is the electric field before the Teflon is removed? E= V/m
d) What is the charge after the Teflon is removed? Q= C
e) What is the potential difference after the Teflon is removed? V= V
f) What are the electric field after the Teflon is removed?
To solve this problem, we can use the formulas related to electric charge, potential difference, and electric field. The first step is to find the area of the electrodes, and then we can use this information to calculate the required values. Let's go step by step:
a) What is the charge before the Teflon is removed?
The formula for the charge (Q) on a capacitor is given by Q = CV, where C is the capacitance and V is the potential difference.
Given that the electrodes have a diameter of 2.5 cm, we can calculate the radius (r) using the formula r = diameter/2:
r = 2.5 cm / 2 = 1.25 cm = 0.0125 m.
The capacitance (C) of a parallel plate 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.
Since the electrodes have a diameter of 2.5 cm, the area (A) of each electrode is given by the formula A = πr²:
A = π(0.0125 m)² = 0.0004908 m².
Given that the thickness (d) of the Teflon sheet is 0.10 mm, we convert it to meters:
d = 0.10 mm = 0.0001 m.
Now, we can calculate the capacitance (C):
C = ε₀A/d.
The value of ε₀ is approximately 8.85 x 10⁻¹² F/m:
C = (8.85 x 10⁻¹² F/m)(0.0004908 m²)/(0.0001 m) ≈ 4.59 x 10⁻¹² F.
Finally, we can calculate the charge (Q) using the formula Q = CV:
Q = (4.59 x 10⁻¹² F)(9.0 V) ≈ 4.13 x 10⁻¹¹ C.
Therefore, the charge before the Teflon is removed is approximately 4.13 x 10⁻¹¹ C.
b) What is the potential difference before the Teflon is removed?
The potential difference (V) is given as 9.0 V in the question.
Therefore, the potential difference before the Teflon is removed is 9.0 V.
c) What is the electric field before the Teflon is removed?
The electric field (E) between the electrodes is given by the formula E = V/d.
Given that the potential difference (V) before the Teflon is removed is 9.0 V, and the distance (d) between the plates is 0.0001 m, we can calculate the electric field:
E = 9.0 V / 0.0001 m.
E = 9.0 x 10⁴ V/m.
Therefore, the electric field before the Teflon is removed is approximately 9.0 x 10⁴ V/m.
d) What is the charge after the Teflon is removed?
When the Teflon is removed, the charge on the electrodes will remain the same, as the battery is still connected. Therefore, the charge after the Teflon is removed is also approximately 4.13 x 10⁻¹¹ C.
e) What is the potential difference after the Teflon is removed?
Since the charge and the capacitance remain the same after the Teflon is removed, the potential difference will also remain the same. Therefore, the potential difference after the Teflon is removed is 9.0 V.
f) What is the electric field after the Teflon is removed?
Again, since the potential difference and the distance between the plates remain the same, the electric field after the Teflon is removed will also be the same. Therefore, the electric field after the Teflon is removed is also approximately 9.0 x 10⁴ V/m.
To answer these questions, we need to understand the properties of the system before and after the Teflon is removed. Let's break down each question one by one.
a) To find the charge before the Teflon is removed, we need to use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. The capacitance of a parallel-plate capacitor is given by C = (ε₀A) / d, where ε₀ is the permittivity of free space, A is the area of the electrodes, and d is the separation between the electrodes. In this case, we have two electrodes with a diameter of 2.5 cm each, so the radius is 1.25 cm or 0.0125 m. The area of one electrode is A = πr², so A = π(0.0125 m)². The separation between the electrodes is the thickness of the Teflon, which is 0.10 mm or 0.00010 m. Now we can calculate the capacitance C using C = (ε₀A) / d. The permittivity of free space ε₀ is approximately 8.85 x 10^-12 F/m. Plug in the values and solve for C.
b) To find the potential difference before the Teflon is removed, we can use the given voltage of the battery, which is 9.0V. The potential difference between the electrodes is equal to the voltage of the battery.
c) To find the electric field before the Teflon is removed, we can use the formula E = V / d, where E is the electric field, V is the potential difference, and d is the separation between the electrodes.
d) To find the charge after the Teflon is removed, we need to consider that removing the Teflon changes the capacitance of the system. After removing the Teflon, the separation between the electrodes is now the diameter of the electrodes, which is 2.5 cm or 0.025 m. We can calculate the new capacitance C' using the formula C' = (ε₀A) / d', where A is the area of the electrodes and d' is the new separation. Plug in the values and solve for C'. Then, use the formula Q' = C'V to find the new charge.
e) To find the potential difference after the Teflon is removed, we can still use the given voltage of the battery, which remains 9.0V. The potential difference between the electrodes is equal to the voltage of the battery.
f) To find the electric field after the Teflon is removed, we can use the formula E' = V / d', where E' is the electric field, V is the potential difference, and d' is the separation between the electrodes.
Now, you can follow these steps and perform the calculations to find the answers to each question.
a) What is the charge before the Teflon is removed? Q= C
NO
Q = C V
same in part d
The field E is 9 Volts/(10^-4 mters)
with or without teflon
However C depends on the material