Question

17. A parallel plate capacitor has a capacity of 20 uF. What will be it's new capacity if
i) the distance between two plates is doubled.
ii) a marble slab of dielectric constant 8 is introduced filling entire space between the
two plates.​

Answer 1

Answer:

hope it will help you

Explanation:

Initial capacitance of capacitor is given by:

C=

d

ε

0

A

20μF=

d

ε

0

A

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

C

1

=

d

kε

0

A

C

1

=kC

C

1

=8×20

C

1

=160μF

After the introduction of dielectric the capacitance increases to 160μF.