Question

An air filled cylindrical capacitor (capacitance \( C_0 \)) of length \( L \), with \( a \) and \( b \) as its inner and outer radii, respectively, consists of two coaxial conducting surfaces. Its cross-sectional view is shown in Fig. (i). In order to increase the capacitance, a dielectric material of relative permittivity \( \varepsilon_r \) is inserted inside 50% of the annular region as shown in Fig. (ii). The value of \( \varepsilon_r \) for which the capacitance of the capacitor in Fig. (ii), becomes \( 5C_0 \) is 

• A. 4
• B. 5
• C. 9
• D. 10

Hint:

When a capacitor is partially filled with dielectric in **parallel regions**, total capacitance is the sum of individual capacitances. Use: \[ C = \frac{1}{2} C_{{air}} + \frac{1}{2} C_{{dielectric}} = C_0 \cdot \frac{1 + \varepsilon_r}{2} \] Equating this with the given new capacitance allows solving for \( \varepsilon_r \).

Answer 1

The Correct Option is C

The original capacitance of a cylindrical capacitor with air:

\[ C_0 = \frac{2\pi \varepsilon_0 L}{\ln(b/a)} \]

In Fig. (ii), the space is equally divided:
- 50% with permittivity \( \varepsilon_0 \)
- 50% with permittivity \( \varepsilon_0 \varepsilon_r \)

Since these two regions are in parallel (same potential difference), the total capacitance is:

\[ C = \frac{1}{2} \cdot \frac{2\pi \varepsilon_0 L}{\ln(b/a)} + \frac{1}{2} \cdot \frac{2\pi \varepsilon_0 \varepsilon_r L}{\ln(b/a)} = \frac{2\pi \varepsilon_0 L}{\ln(b/a)} \cdot \frac{1 + \varepsilon_r}{2} \]

But:

\[ C_0 = \frac{2\pi \varepsilon_0 L}{\ln(b/a)} \Rightarrow C = C_0 \cdot \frac{1 + \varepsilon_r}{2} \]

Given:

\[ C = 5C_0 \Rightarrow C_0 \cdot \frac{1 + \varepsilon_r}{2} = 5C_0 \Rightarrow \frac{1 + \varepsilon_r}{2} = 5 \Rightarrow 1 + \varepsilon_r = 10 \Rightarrow \varepsilon_r = 9 \]

\[ \boxed{\varepsilon_r = 9} \]