Question

A parallel-plate capacitor has plates with area 2.90×10^−2 m^2 separated by 1.60 mm of Teflon.

A) Calculate the charge on the plates when they are charged to a potential difference of 10.0 VV .
B) Use Gauss's law to calculate the electric field inside the Teflon.
C) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

Answer 1

A) The charge on the plates when charged to a potential difference of 10.0 V is approximately 5.10x10⁻¹⁰ C. B) The electric field inside the Teflon is zero. C) The electric field when the Teflon is removed is approximately 6.25x10³ V/m.

Step-by-step explanation:

A) To calculate the charge on the plates, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. The capacitance can be calculated using 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. Plugging in the given values, we get C = (8.85x10⁻¹² F/m)(2.90x10⁻² m²)/(1.60x10⁻³ m) ≈ 5.10x10⁻¹¹ F. Then, calculating the charge, Q = (5.10x10⁻¹¹ F)(10.0 V) ≈ 5.10x10⁻¹⁰ C.

B) Gauss's law states that the electric field inside a conductor is zero. Since the capacitor is filled with Teflon, which is a dielectric, the electric field inside the Teflon is also zero.

C) When the Teflon is removed, the electric field inside the capacitor is given by E = V/d, where E is the electric field, V is the potential difference, and d is the distance between the plates. Plugging in the given values, we get E = (10.0 V)/(1.60x10⁻³ m) = 6.25x10³ V/m.

Answer 2

To calculate the charge on the plates of the parallel-plate capacitor, use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. The electric field can be calculated using Gauss's law, and when the Teflon is removed, the electric field is zero.

Step-by-step explanation:

To calculate the charge on the plates of the parallel-plate capacitor when they are charged to a potential difference of 10.0 V, we can 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 = εA/d, where ε is the permittivity of the dielectric material, A is the area of the plates, and d is the separation between the plates. In this case, the capacitance can be calculated as C = (ε0εrA)/d, where ε0 is the permittivity of free space (8.85 x 10^-12 F/m), εr is the relative permittivity of Teflon (2.1), A is the given area (2.90 x 10^-2 m^2), and d is the given separation (1.60 mm).

Using these values, we can calculate the capacitance. The charge on the plates can then be found by multiplying the capacitance by the potential difference.

To calculate the electric field inside the Teflon using Gauss's law, we can use the formula E = σ/εrε0, where σ is the surface charge density, εr is the relative permittivity of Teflon, and ε0 is the permittivity of free space. The surface charge density can be found by dividing the charge on the plates by the area of the plates.

When the Teflon is removed and the voltage source is disconnected, there is no charge on the plates. Therefore, the electric field is zero in this case.