Question

If we add a dielectric material with a dielectric constant k=3.0 between the plates of a fully-charged 2.0μF capacity while a 2 kV supply remains connected (ie, V constant, Q can change), by what factor does it affect the following (ie, is it the same, double, halved, etc)?
i) the electric field,
ii) the capacitance,
iii) the potential energy energy stored in the capacity.
Be sure to justify your answer using the appropriate equation.

Answer 1

1. We have : $V = E\times d$, where d is a distance between the plates of a capacitor. As both $V,d$ don't change, $E$ remains the same.
2. The capacitance of a parallel-plate capacitor is given by $C=\frac{k \varepsilon_0 S}{d}$ , so the capacitance will triple (as $k_{air}=1$ )
3. The energy is given by $W=\frac{VQ}{2}=\frac{CV^2}{2}$ will be tripled, as $V$ stays the same and $C$ triples.