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3 years ago

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Two large parallel conducting plates are 8.0 cm apart and carry equal but opposite charges on their facing surfaces. The magnitu

de of the surface charge density on either of the facing surfaces is 2.0 nc/m2. Determine the magnitude of the electric potential difference between the plates.

1 answer:

Pepsi [2]3 years ago

8 0

Answer:

18.1 V

Explanation:

The electric field between two parallel plates is given by the equation:

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

where

$\sigma$ is the charge surface density

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

For the plates in this problem,

$\sigma = 2.0 nC/m^2 = 2.0\cdot 10^{-9} C/m^2$

So, the magnitude of the electric field is

$E=\frac{2.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=226.0 V/m$

Now we can find the potential difference between the plates, which is given by

$\Delta V = E d$

where

d = 8.0 cm = 0.08 m is the separation between the plates

Substituting,

$\Delta V=(226.0)(0.08)=18.1 V$

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