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

`1.78\cdot 10^(-4) m`

Step-by-step explanation:

The capacitance of the capacitor is given by:

`C=(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=(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(\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=(k\epsilon_0 A)/(C)=((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`