A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is 2.40 10-3 m, and that of the outer shell is 2.53 10-3 m. When the cylinders carry equal and opposite charges of magnitude 1.53 10-10 C, the electric field between the plates has an average magnitude of 3.85 104 V/m and is directed radially outward from the inner shell to the outer shell.

(a) Determine the magnitude of the potential difference between the cylindrical shells.

(b) Determine the capacitance of this capacitor.

(a) The magnitude of the potential difference between the cylindrical shells can be calculated using the formula V = Ed, where V is the potential difference, E is the electric field, and d is the distance between the shells. In this case, the distance between the shells is the difference in their radii.

Let's calculate the distance:

d = (2.53 × 10^-3) m - (2.40 × 10^-3) m
= (0.13 × 10^-3) m
= 1.3 × 10^-4 m

Now, we can calculate the potential difference:

V = (3.85 × 10^4) V/m × (1.3 × 10^-4) m
= 5.005 V

Therefore, the magnitude of the potential difference between the cylindrical shells is 5.005 V.

(b) The capacitance of a capacitor can be calculated using the formula C = Q / V, where C is the capacitance, Q is the charge, and V is the potential difference.

Let's calculate the capacitance:

C = (1.53 × 10^-10) C / (5.005 V)
= 3.057 × 10^-11 F

Therefore, the capacitance of this capacitor is 3.057 × 10^-11 F.

To solve this problem, we can use the formula for the electric field between the plates of a capacitor:

E = V / d

where E is the electric field magnitude, V is the potential difference between the plates, and d is the distance between the plates.

(a) To find the potential difference between the cylindrical shells, we rearrange the formula as:

V = E * d

Given that the electric field magnitude E is 3.85 x 10^4 V/m and the distance between the shells (d) is the difference in their radii, we can calculate V as:

V = (3.85 x 10^4 V/m) * (2.53 x 10^-3 m - 2.40 x 10^-3 m)
= (3.85 x 10^4 V/m) * (0.13 x 10^-3 m)
= 5.005 x 10^-1 V

The magnitude of the potential difference between the cylindrical shells is approximately 0.501 V.

(b) The capacitance of a capacitor is given by the formula:

C = Q / V

where C is the capacitance, Q is the magnitude of the charge on either cylindrical shell, and V is the potential difference between the shells.

Given that the charge Q is 1.53 x 10^-10 C and the potential difference V is 0.501 V, we can calculate C as:

C = (1.53 x 10^-10 C) / (0.501 V)
= 3.056 x 10^-10 F

The capacitance of this capacitor is approximately 3.056 x 10^-10 Farads.

To determine the magnitude of the potential difference between the cylindrical shells, you need to use the formula for electric field, which is given by:

E = V/d

where E is the electric field magnitude, V is the potential difference, and d is the separation between the shells.

Given that the electric field (E) is 3.85 x 10^4 V/m, we can rearrange the formula to solve for V:

V = E x d

The separation between the shells (d) is the difference between the radii of the outer and inner shells:

d = r_outer - r_inner = 2.53 x 10^-3 m - 2.40 x 10^-3 m

Now you can calculate the potential difference (V) by substituting the values into the equation:

V = (3.85 x 10^4 V/m) x (2.53 x 10^-3 m - 2.40 x 10^-3 m)

Solving this equation will give you the magnitude of the potential difference between the cylindrical shells.

To determine the capacitance of the capacitor, you can use the formula:

C = Q / V

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

Given that the magnitude of the charge (Q) is 1.53 x 10^-10 C (which is the same for both shells), and you already calculated the potential difference (V) in the previous step, you can simply substitute these values into the equation to solve for capacitance (C).

C=Q/V figure voltage as E*distance, where distance is the difference of the radii.