[tex]
\mathbb \blue{Answer }[/tex]
[tex]Introducing \: \: a \: dielectric \: \: into \: \\ a \: \: capacitor \: \: decreases \: \: \\ the \: \: electric \: \: field, \: \: \\ which \: \: decreases \: \: the \: \: \\ voltage, \: \: \ which \: \: increases \: \: the \: \: \\ capacitance. \: \: A \: capacitor \: \: with \: a \: \\ dielectric \: \: stores \: the \: \\ same \: \: \ charge \: \: as \: one \: \\ without \: \: a \: dielectric, \: but \: at \: \\ a \: \: lower \: \: voltage. \: \: \\ Therefore \: \: a \: \: capacitor \: \: with \: \: a \: \\ dielectric \: \: in \: \: it \: \\ is \: \: more \: \: effective \: [/tex]
[tex]\mathbb \blue{ I \: hope \: it's \: helpful \: \: for \: you }[/tex]
,,,,,,,,,,,,,,,,,,,,,,,,,
