To solve this problem, we need to determine the relative permittivity \epsilon_r such that the capacitance of the modified cylindrical capacitor (Figure (ii)) becomes 5C_0.

The original air-filled cylindrical capacitor (Figure (i)) has a capacitance given by the formula:

C_0 = \frac{2\pi\epsilon_0L}{\ln(b/a)}

where:

• L = length of the cylinder
• a = inner radius
• b = outer radius
• \epsilon_0 = permittivity of free space

In the modified capacitor (Figure (ii)), the dielectric material with relative permittivity \epsilon_r is inserted into 50% of the annular region. Therefore, the annular region is divided into two parts:

• Half with dielectric \epsilon_0\epsilon_r
• Half remaining with dielectric \epsilon_0

The capacitance C of this configuration can be considered as two capacitors in parallel:

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}

where:

• C_1 = \frac{\pi\epsilon_0L}{\ln(b/a)} for the air part
• C_2 = \frac{\pi\epsilon_0\epsilon_rL}{\ln(b/a)} for the dielectric part

We need both capacitors because the dielectric is inserted only in half the region:

\frac{1}{C} = \frac{1}{\frac{\pi\epsilon_0L}{\ln(b/a)}} + \frac{1}{\frac{\pi\epsilon_0\epsilon_rL}{\ln(b/a)}}

C = \frac{\pi\epsilon_0L}{\ln(b/a)} \frac{\epsilon_r}{1 + (\epsilon_r - 1)/2}

For the capacitance C to be 5C_0, we set:

\frac{\pi\epsilon_0L}{\ln(b/a)} \frac{\epsilon_r}{(1 + (\epsilon_r - 1)/2)} = 5 \frac{2\pi\epsilon_0L}{\ln(b/a)}

Solving, we simplify:

\frac{\epsilon_r}{1 + \frac{\epsilon_r - 1}{2}} = 10

\frac{2\epsilon_r}{\epsilon_r + 1} = 10

2\epsilon_r = 10(\epsilon_r + 1)

2\epsilon_r = 10\epsilon_r + 10

\epsilon_r = \frac{10}{8} = 1.25

However, on recalculating, there seems to be an error, as the capacitance relation was miswritten earlier. Correcting:

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