A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about $10^{7}\ \text{V m}^{-1}$. For safety we would like the field never to exceed, say, 10\% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

$A \approx 19\ \text{cm}^2$ ($\approx 18.8 \times 10^{-4}\ \text{m}^2$).

CBSE Class 12 · 2005 · All India · CapacitanceShort Answer

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about

1 0^{7} V m^{- 1}

. For safety we would like the field never to exceed, say, 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Answer: $A \approx 19 cm^{2}$ ( $\approx 18.8 \times 1 0^{- 4} m^{2}$ ).

Solution

Maximum permissible field

= 10%

1 0^{7}

E = 1 0^{6} V m^{- 1}

For a rating

V = 1 0^{3} V

, the minimum plate separation is:

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

From

C = \frac{K ε _{0} A}{d}

A = \frac{C d}{K ε _{0}} = \frac{50 \times 1 0 ^{- 12} \times 1 0 ^{- 3}}{3 \times 8.85 \times 1 0 ^{- 12}}

A = 18.8 \times 1 0^{- 4} m^{2} \approx 19 cm^{2}

CBSE Class 12 Board · 2005 · All India · Subjective

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