Question

Dielectrics play an important role in design of capacitors. The molecules of a dielectric may be polar or non-polar. When a dielectric slab is placed in an external electric field, opposite charges appear on the two surfaces of the slab perpendicular to electric field. Due to this an electric field is established inside the dielectric. The capacitance of a capacitor is determined by the dielectric constant of the material that fills the space between the plates. Consequently, the energy storage capacity of a capacitor is also affected. Like resistors, capacitors can also be arranged in series and/or parallel.

An air-filled capacitor with plate area A and plate separation d has capacitance C₀. A slab of dielectric constant K, area A and thickness `(d/5)` is inserted between the plates. The capacitance of the capacitor will become ______.

पर्याय

• `[(4K)/(5K + 1)]C_0`

• `[(K + 5)/(4)]C_0`

• `[(5K)/(4K + 1)]C_0`

• `[(K + 4)/(5K)]C_0`

Answer 1

An air-filled capacitor with plate area A and plate separation d has capacitance C₀. A slab of dielectric constant K, area A and thickness `(d/5)` is inserted between the plates. The capacitance of the capacitor will become `bbunderline([(5K)/(4K + 1)]C_0)`.

Explanation:

Since, `C_0 = (Aepsilon_0)/d`

Now, `C = (Aepsilon_0)/(d - t + t/k)`

`C = (Aepsilon_0)/(d - d/5 + d/(5k))   ...[because t = d/5]`

`C = (Aepsilon_0)/d[1/(1 - 1/5 + 1/(5K))]`

`C = [(5K)/(1 + 4K)]C_0`