Dielectrics

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.