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3 years ago

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Ask Your Teacher When a potential difference of 160 V is applied to the plates of a parallel-plate capacitor, the plates carry a

surface charge density of 26.0 nC/cm2. What is the spacing between the plates?

1 answer:

Morgarella [4.7K]3 years ago

7 0

The spacing between the plates is $5.4\cdot 10^{-6} m$

Explanation:

The electric field in the spacing between the plates of a parallel-plate capacitor is uniform and it is given by

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

We also know that the relationship between electric field and potential difference across a parallel-plate capacitor is

$V=Ed$

where

V is the potential difference

d is the spacing between the plates

Combining the two equations we get

$V=\frac{\sigma d}{\epsilon_0}$

And given that here we have:

V = 160 V

$\sigma = 26.0 nC/cm^2 = 26.0\cdot 10^{-9} C/cm^2=26.0\cdot 10^{-5} C/m^2$

We can solve the equation for d, the spacing between the plates:

$d=\frac{V \epsilon_0}{\sigma}=\frac{(160)(8.85\cdot 10^{-12})}{26.0\cdot 10^{-5}}=5.4\cdot 10^{-6} m$

Learn more about capacitors:

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