A parallel plate capacitor is to be designed with a voltage rating 1 kV using a material of dielectric constant 10 and dielectric strength 106 V / m -1. What minimum area of the plates (in 10-3 m 2 ) is required to have a capacitance of 88.5 pF ?

Question

A parallel plate capacitor is to be designed with a voltage rating 1 kV using a material of dielectric constant 10 and dielectric strength 106 V / m -1. What minimum area of the plates (in 10-3 m 2 ) is required to have a capacitance of 88.5 pF ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For dielectric strength maximum potential difference which can be applied across capacitor is related as

E_{B} = \frac{V}{d} \Rightarrow d = \frac{V}{E} = \frac{1 0 ^{3}}{1 0 ^{6}} = 1 0^{- 3} m

We use capacitance of capacitor

C = \frac{\in _{0} \in _{r} A}{d}

\Rightarrow A = \frac{C d}{ϵ _{0} \in _{r}}

\Rightarrow A = \frac{88.5 \times 1 0 ^{- 12} \times 1 0 ^{- 3}}{8.85 \times 1 0 ^{- 12} \times 10} = 1 0^{- 3} m^{2}

Q. A parallel plate capacitor is to be designed with a voltage rating 1kV using a material of dielectric constant 10 and dielectric strength 106V/m−1. What minimum area of the plates (in 10−3m2 ) is required to have a capacitance of 88.5pF ?

For dielectric strength maximum potential difference which can be applied across capacitor is related as EB​=dV​⇒d=EV​=106103​=10−3m We use capacitance of capacitor C=d∈0​∈r​A​ ⇒A=ϵ0​∈r​Cd​ ⇒A=8.85×10−12×1088.5×10−12×10−3​=10−3m2

Q. A parallel plate capacitor is to be designed with a voltage rating $1 kV$ using a material of dielectric constant $10$ and dielectric strength $1 0^{6} V / m^{- 1}$ . What minimum area of the plates (in $1 0^{- 3} m^{2}$ ) is required to have a capacitance of $88.5 pF$ ?