Which Statement Most Accurately Captures What Current Evidence Tell Us About The Habitability Of Mars?

a) The capacitance of the ceramic capacitor is approximately 354.16 pF.

b) The potential difference between the plates of the capacitor will be approximately 3.39 x 10^6 volts.

To calculate the capacitance of the ceramic capacitor, we can use the formula:

C = (ε₀ * εᵣ * A) / d

where:

C is the capacitance,

ε₀ is the vacuum permittivity (approximately 8.854 x 10^(-12) F/m),

εᵣ is the relative permittivity of the ceramic (given as 100),

A is the effective plate area (given as 4 cm², which is equal to 4 x 10^(-4) m²),

d is the separation between the plates (given as 0.1 mm, which is equal to 0.1 x 10^(-3) m).

Let’s calculate the capacitance in picofarads (pF):

a) Calculation of capacitance (C):

C = (ε₀ * εᵣ * A) / d

= (8.854 x 10^(-12) F/m * 100 * 4 x 10^(-4) m²) / (0.1 x 10^(-3) m)

= (8.854 x 100 x 4) / 0.1

= 354.16 pF

Therefore, the capacitance of the ceramic capacitor is approximately 354.16 pF.

b) To find the potential difference (PD) between the plates when the capacitor is given a charge of 1.2 μC (microcoulombs), we can use the formula:

PD = Q / C

where:

PD is the potential difference,

Q is the charge (given as 1.2 μC, which is equal to 1.2 x 10^(-6) C),

C is the capacitance (calculated in part a) as 354.16 pF, which is equal to 354.16 x 10^(-12) F).

Let’s calculate the potential difference (PD):

b) Calculation of potential difference (PD):

PD = Q / C

= (1.2 x 10^(-6) C) / (354.16 x 10^(-12) F)

= 3.39 x 10^6 V

Therefore, the potential difference between the plates of the capacitor will be approximately 3.39 x 10^6 volts.

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