Question

Two parallel plates which have a cross-sectional area of 100 cm², carry an equal
and opposite charge of 1.0×10^-7 C. The space between the plates is filled with a dielectric material and the electric field within the dielectric is 2.3 x 10^5 Vm^-1. If the electric field across the plates without the dielectric is given by E0 = sigma/aphselonot sigma being the surface charge density of the plates, calculate the dielectric constant of the dielectric.

Answer 1

The dielectric constant can be obtained as a ratio between the corresponding values of the electric field (without dielectric material and when dielectric material is inserted):

$$\varepsilon = \frac{E_0}{E_\varepsilon},$$

where

$$E_0 = \frac{\sigma}{\varepsilon_0}$$

The surface charge density can be obtained as follows:

$$\sigma = \frac{q}{S}$$

Substituting all values into the initial expression, we obtain:

$$\varepsilon = \frac{q}{\varepsilon_0 S E_{\varepsilon}}$$

Finally,

$$\varepsilon = \frac{10^{-7}}{8.85 \cdot 10^{-12} \cdot 10^{-2} \cdot 2.3 \cdot 10^{5}} \approx 4.9$$

Answer: 4.9