Dielectrics

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A parallel-plate capacitor with plate separation 4 mm is fully filled with a dielectric of relative permittivity $\varepsilon_r = 5$. By what factor does the capacitance change compared to the vacuum-filled capacitor?

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When a dielectric material fully fills the space between the plates of a capacitor, the capacitance increases by the relative permittivity (dielectric constant) $\varepsilon_r$ of the material. This occurs because the dielectric reduces the electric field inside it by polarizing: the bound charges in the dielectric partially cancel the free surface charges on the plates, reducing the potential difference for the same stored charge, and thus increasing $C = Q/V$. Mathematically, $C = \varepsilon_r \varepsilon_0 A/d$, so the capacitance becomes $\varepsilon_r$ times the vacuum capacitance. With $\varepsilon_r = 5$, the new capacitance is exactly 5 times the vacuum value. Option 'decreases by factor 5' is incorrect because a dielectric always increases capacitance, never decreases it. Option 'increases by factor 25' is incorrect because it confuses $\varepsilon_r$ with $\varepsilon_r^2$. Option 'remains unchanged' would only hold if the dielectric constant were 1, i.e., vacuum. The plate separation of 4 mm is given as context but is not needed for computing the factor of change. Plausibility check: all real dielectrics have $\varepsilon_r > 1$, so filling with any dielectric must increase capacitance.

Dielectrics

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