Question

The capacitance of a parallel plate capacitor is increased by a factor of 7 when a dielectric
material fills the space between its plates. What is the relative permitivity of the dielectric
material? If this material is placed in between the plates of a cylindrical capacitor of outer
and inner radii 12 cm and 10 cm, respectively, calculate the capacitance per unit length of
the cylindrical capacitor.

Answer 1

For a flat capacitor, we write

$C_1=\frac{\epsilon_1 \cdot \epsilon_0 \cdot S}{d}$

$C_2=\frac{\epsilon_2 \cdot \epsilon_0 \cdot S}{d}$

Where $\epsilon_1=1$ $C_2=7 \cdot C_1$

Then write

$\epsilon_1=\frac{C_1d }{\epsilon_0 \cdot S}$

$\epsilon_2=\frac{C_2d }{\epsilon_0 \cdot S}$

$\frac{\epsilon_2 }{\epsilon_1 }=\frac{\frac{C_2d }{\epsilon_0 \cdot S}}{\frac{C_1d }{\epsilon_0 \cdot S} }=\frac{C_2 }{C_1 }=\frac{7C_1 }{C_1 }=7$

Then $\epsilon =7$

Material: mica or glass

For a cylindrical capacitor, we write

$C=\frac{2\pi\epsilon_0\epsilon l}{ln(r_2/r_1) }=\frac{2\cdot 3.14 \cdot 8.85 \cdot 10^{-12} \cdot 7 \cdot 1 }{ln(0.12/0.1) }=2.135 \cdot 10^{-9} F=2.135 nF$