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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 |
the electric field after inserting the slab is $\frac{E_0}{k}$ the capacitance after inserting the slab is $k C_0$ the induced charge on the slab is $\mathrm{Q}\left(1-\frac{1}{k}\right)$ All of the above |
All of the above |
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. $q'=q\left(1-\frac{1}{K}\right)$. |
