Ncert Exercises & Solutions

A parallel plate capacitor is to be designed with a voltage rating \(1~\text{kV},\) using a material of dielectric constant \(3\) and dielectric strength about \(10^7~\text {Vm}^{-1}.\) (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say \(10\text %\) of the dielectric strength. What minimum area of the plates is required to have a capacitance of \(50~\text{pF}?\)
1. \(17~\text{cm}^2\)
2. \(21~\text{cm}^2\)
3. \(15~\text{cm}^2\)
4. \(19~\text{cm}^2\)

Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V

The dielectric constant of a material,

\in_{r}

= 3

Dielectric strength of 10⁷ V/m

For safety, the field intensity never exceeds 10% of the dielectric strength.

Hence, electric field intensity, E = 10% of 10⁷ = 10⁶ V/m

The capacitance of the parallel plate capacitor, C = 50 pF = 50 x 10^-12 F Distance between the plates is given by,

\begin{aligned} d & = \frac{V}{E} \\ = \frac{1000}{10^{6}} = 10^{- 3} m \end{aligned}

Capacitance is given by the relation,

C = \frac{ϵ_{0} ϵ_{r} A}{d}

Where,
A = Area of each plate

\begin{aligned} ϵ_{0} & = Permittivity of free space = 8 .85 \times 10^{- 12} N^{- 1} C^{2} m^{- 2} \\ ∴ A & = \frac{Cd}{ϵ_{0} \in_{r}} \\ = \frac{50 \times 10^{- 12} \times 10^{- 3}}{8 .85 \times 10^{- 12} \times 3} \approx 19 {cm}^{2} \end{aligned}

Hence, the area of each plate is about 19 cm².

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