Two large parallel plates are separated by $1.0 \mathrm {~mm}$ of air. If the potential difference between them is $3.0 \mathrm {~V}$, what is the magnitude of their surface charge densities? $\left( \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right)$

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Quizplus · Accepted Answer

The electric field $E$ between two parallel plates is given by $E = \frac{V}{d}$, where $V$ is the potential difference and $d$ is the separation between the plates. The surface charge density $\sigma$ is related to the electric field by $\sigma = \varepsilon_0 E$. Substituting $E = \frac{V}{d}$ into $\sigma = \varepsilon_0 E$, we get $\sigma = \varepsilon_0 \frac{V}{d}$. Plugging in the given values, $\sigma = 8.85 \times 10^{-12} \frac{3.0}{1.0 \times 10^{-3}} = 2.655 \times 10^{-8} \mathrm{C/m^2}$, which rounds to $2.7 \times 10^{-8} \mathrm{C/m^2}$.

Two Large Parallel Plates Are Separated By 1.0 mm1.0 \mathrm {~mm}1.0 mm Of Air If the Potential Difference Between Them Is

Two Large Parallel Plates Are Separated By $1.0 \mathrm {~mm}$ Of Air If the Potential Difference Between Them Is