A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is 2.35 10-3 m, and that of the outer shell is 2.49 10-3 m. When the cylinders carry equal and opposite charges of magnitude 1.61 10-10 C, the electric field between the plates has an average magnitude of 3.80 104 V/m and is directed radially outward from the inner shell to the outer shell.

Determine the magnitude of the potential difference between the cylindrical shells.
Determine the capacitance of this capacitor.

You will find the formulas that you need here:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capcyl.html

In this case, the gap is so narrow compared to the radius that you can get very high accuracy by treating it as a planar capacitor with gap 0.14*10^-3 m. Multiply that by the average E-field between the cylindrical plates, to get the potential difference, 5.32 V .

The capacitance is C = Q/V

To determine the magnitude of the potential difference between the cylindrical shells, we can use the formula for electric field:

E = V / d

where E is the electric field, V is the potential difference, and d is the separation distance between the shells. We are given that the electric field has an average magnitude of 3.80 x 10^4 V/m.

The separation distance between the shells is the difference in their radii:

d = r_outer - r_inner
= (2.49 x 10^-3 m) - (2.35 x 10^-3 m)
= 0.14 x 10^-3 m

Now we can rearrange the formula for electric field to solve for V:

V = E * d
= (3.80 x 10^4 V/m) * (0.14 x 10^-3 m)
= 5.32 V

Therefore, the magnitude of the potential difference between the cylindrical shells is 5.32 V.

To determine the capacitance of the capacitor, we 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 is 1.61 x 10^-10 C and the potential difference is 5.32 V, we can calculate the capacitance:

C = (1.61 x 10^-10 C) / (5.32 V)
= 3.03 x 10^-11 F

Therefore, the capacitance of the capacitor is 3.03 x 10^-11 F.

To determine the magnitude of the potential difference between the cylindrical shells, we can use the formula:

ΔV = E * d

where ΔV is the potential difference, E is the electric field strength, and d is the distance between the plates.

In this case, the electric field strength is given as 3.80 x 10^4 V/m. The distance between the plates can be calculated by subtracting the radius of the inner shell from the radius of the outer shell:

d = r_outer - r_inner = (2.49 x 10^-3 m) - (2.35 x 10^-3 m)

Now we can substitute the values into the formula:

ΔV = (3.80 x 10^4 V/m) * ((2.49 x 10^-3 m) - (2.35 x 10^-3 m))

Calculating this expression gives us the magnitude of the potential difference between the cylindrical shells.

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

C = Q / ΔV

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

In this case, the magnitude of the charge is given as 1.61 x 10^-10 C, and we have already calculated the potential difference.

Substituting the values into the formula, we can find the capacitance:

C = (1.61 x 10^-10 C) / ΔV

Calculating this expression gives us the capacitance of the capacitor.