2.33 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 10⁷ 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% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

93 Views | Posted 5 months ago

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

2.33 Potential rating of the capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, $?_{r}$ = 3

Dielectric strength = $10^{7}$ V/m

For safety, the field intensity should not cross 10% of the dielectric strength, hence

Electric field intensity, E = 10 % of $10^{7}$ = $10^{6}$ V/m

Capacitance of the parallel plate capacitor, C = 50 pF = 50 $\times 10^{- 12}$ F

Distance between the plates, d is given by d = $\frac{V}{E}$ = $\frac{1000}{10^{6}}$ m = $10^{- 3}$ m

Capacitance is given by the relation

C = $\frac{ε_{0} ?_{r} A}{d}$ , where $ε_{0}$ = permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

50 $\times 10^{- 12}$ = $\frac{8.854 \times 10^{- 12} \times 3 \times A}{10^{- 3}}$

A = 1.88 $\times 10^{- 3}$ $m^{2}$ = 18.8 ${c m}^{2}$

Hence, t

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