A parallel plate capacitor with plate area A and separation d has charge Q. A slab of dielectric constant k is inserted in space between the plates almost completely fills the space. If E0 and C0 be the electric field and capacitance before inserting the slab, then
Answer & explanation
Correct answer: option 4
Due to induced charges on dielectric electric field reduces to $\frac{E_0}{K}$ and hence potential difference decreases in same proportion. This results in increase in capacitances by K times.
$q'=q\left(1-\frac{1}{K}\right)$.