Question

a slab of material of dielectric constant K has the same area as the plates of parallel plate capacitor but has a thickness 3/4 D where D is separation between plates how is the capacitance change when the slab is inserted between plates ​

Answer 1

Answer:

Increasing by a factor of 16k/3

Explanation:

Capacitance, initially is given by C=AĖ/D where Ė permitivitty of free space and D is separation between plates

Since there is a gap of distance d/4 with air as the medium between the capacitors and the remaining volume is filled by dielectric, you can consider it as two capacitance of capacitance C1 and C2 respectively, connected in series

Where

C1= AĖ/(d/4) = 4AĖ/d

C2= KAĖ(3d/4) = 4KAĖ/3d

Effective capacitance =4KAĖ/(3+k)d

Ratio of increment = 4K/(3+k)

Please refer to the attachment

Hope this answer helped you