Question

When the electric field is reduced to \[E'=\frac{E}{K}\] by a dielectric, what happens to potential difference and capacitance?

Options

• \[V'=E'\cdot d\] decreases, and capacitance decreases by factor \[K\]

• \[V'=E'\cdot d\] stays unchanged, and capacitance stays unchanged

• \[V'=E'\cdot d\] decreases, and capacitance increases by factor \[K\]

• \[V'=E'\cdot d\] increases, and capacitance decreases by factor \[K\]

Answer 1

The reduced electric field makes \[V'=E'\cdot d\] decrease. Since \[C=\frac{Q}{V'}\], the capacitance increases by factor \[K\].