Question

The effective area of each plate of a parallel plate capacitor is 2.1 m2. The capacitor is filled with neoprene rubber (κ = 6.4). When a 6.0-V potential difference exists across the plates of the capacitor, the capacitor stores 4.0 µC of charge. Determine the plate separation of the capacitor.

Answer 1

Answer:

$1.78\cdot 10^{-4} m$

Explanation:

The capacitance of the capacitor is given by:

$C=\frac{Q}{V}$

where

$Q=4.0\mu C=4.0\cdot 10^{-6} C$ is the charge stored on the plates

V = 6.0 V is the potential difference across the capacitor

Substituting, we find

$C=\frac{4.0\cdot 10^{-6} C}{6.0 V}=6.7\cdot 10^{-7} F$

The capacitance of a parallel-plate capacitor is also given by

$C=k\frac{\epsilon_0 A}{d}$

where

k = 6.4 is the dielectric constant

$\epsilon_0$ is the vacuum permittivity

$A=2.1 m^2$ is the area of each plate

d is the separation between the plates

Solving for d, we find

$d=\frac{k\epsilon_0 A}{C}=\frac{(6.4)(8.85\cdot 10^{-12} F/m)(2.1 m^2)}{6.7\cdot 10^{-7}F}=1.78\cdot 10^{-4} m$