A capacitor consists of two concentric spherical shells. The outer radius of the inner shell is a=0.43 mm and the inner radius of the outer shell is b=2.2 mm .

(a) What is the capacitance C of this capacitor? Express your answer in Farads.

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(b) Suppose the Maximum possible electric field at the outer surface of the inner shell before the air starts to ionize is Emax(a)=3.0*10^6V/m. What is the maximum possible charge on the inner sphere? Express your answer in Coulombs.

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(c) What is the maximum amount of energy stored in the capacitor? Express your answer in Joules.

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(d) When Ea)=3.0*10^6V/m what is the absolute value of the potential difference between the shells? Express your answer in Volts.

Please someone answer.

To find the capacitance, maximum charge, maximum energy stored, and potential difference between the shells, we need to use the formulas and properties of a capacitor.

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

C = (4πε₀) * (a*b) / (b - a)

where ε₀ is the permittivity of free space (vacuum), a is the inner radius, and b is the outer radius.

In this case, we have a = 0.43 mm and b = 2.2 mm. First, let's convert these values to meters:

a = 0.43 mm = 0.43 * 10^(-3) m
b = 2.2 mm = 2.2 * 10^(-3) m

Now, we can plug these values into the formula:

C = (4πε₀) * (0.43 * 10^(-3) * 2.2 * 10^(-3)) / (2.2 * 10^(-3) - 0.43 * 10^(-3))

The value of ε₀ is approximately 8.85 * 10^(-12) F/m, so we can substitute it in:

C ≈ (4π * 8.85 * 10^(-12)) * (0.43 * 10^(-3) * 2.2 * 10^(-3)) / (2.2 * 10^(-3) - 0.43 * 10^(-3))

Performing the calculations, we find:

C ≈ 3.00 * 10^(-12) F

Therefore, the capacitance of this capacitor is approximately 3.00 picofarads (pF).

(b) The maximum charge on the inner sphere can be found using the formula:

Q = C * V

where Q is the charge, C is the capacitance, and V is the potential difference.

In this case, we are given the maximum electric field Emax(a) at the outer surface of the inner shell, which is related to the potential difference V by the formula:

Emax(a) = V / b

Rearranging the formula, we can solve for V:

V = Emax(a) * b

Substituting the given values, we have:

V = (3.0 * 10^6 V/m) * (2.2 * 10^(-3) m)

V ≈ 6.6 V

Now, we can calculate the maximum charge:

Q = (3.00 * 10^(-12) F) * (6.6 V)

Q ≈ 1.98 * 10^(-11) C

Therefore, the maximum charge on the inner sphere is approximately 1.98 * 10^(-11) Coulombs.

(c) The maximum amount of energy stored in the capacitor can be calculated using the formula:

U = (1/2) * C * V²

where U is the energy stored, C is the capacitance, and V is the potential difference.

Using the calculated value of the capacitance (C ≈ 3.00 * 10^(-12) F) and the potential difference (V ≈ 6.6 V) obtained earlier, we can substitute them into the formula:

U = (1/2) * (3.00 * 10^(-12) F) * (6.6 V)²

Calculating the expression, we find:

U ≈ 68.94 * 10^(-12) J

Therefore, the maximum amount of energy stored in the capacitor is approximately 68.94 picajoules (pJ).

(d) The potential difference between the shells when the electric field is Emax(a) can be directly obtained by multiplying Emax(a) and the distance between the shells (b - a):

ΔV = Emax(a) * (b - a)

Substituting the given values, we have:

ΔV = (3.0 * 10^6 V/m) * (2.2 * 10^(-3) m - 0.43 * 10^(-3) m)

Performing the calculation, we find:

ΔV ≈ 5.39 V

Therefore, the absolute value of the potential difference between the shells when Emax(a) is 3.0 * 10^6 V/m is approximately 5.39 volts.