Question

A parallel plate air has a capacity of $20\mu F$ . what will be the new capacity if a marble slab of dielectric constant $8$ is introduced between the two plates?

Answer 1

Initial capacitance of capacitor is given by:

$C=\frac{{\epsilon }_{0}A}{d}$

$20\mu F=\frac{{\epsilon }_{0}A}{d}$

Now a dielectric constant is introduced between the capacitor. Now new capacitance is given by:

$C_1=\frac{k{\epsilon }_{0}A}{d}$

$C_1=kC$

$C_1=8\times 20$

$C_1=160\mu F$

After the introduction of dielectric the capacitance increases to $160\mu F$.