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4 years ago

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A capacitor is to be constructed to have a capacitance of 100uF.The area of the plates is 6.om by 0.030m and the relative permit

ivityof dielectric is 7.0 Find the necessary separation of the plates and the electric field strength if a potential difference of 12V is applied across the capacitor.

1 answer:

lara [203]4 years ago

3 0

Answer:

The answer is below

Explanation:

Given that:

The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m = 0.18 m²

the relative permittivity of dielectric (εr) is 7.0

Permittivity of free space (εo) = 8.854 × 10^(-12)

capacitance of 100uF

potential difference (V) of 12V

d = separation between plate

The capacitance (C) of a capacitor is given by:

$C=\frac{\epsilon_o \epsilon_r A }{d}\\ 100*10^{-6}=\frac{8.854*10^{-12}*7*0.18}{d}\\ d=\frac{8.854*10^{-12}*7*0.18}{100*10^{-6}}=1.11*10^{-7}\ m$

The electric field between plates is given as:

E = V /d

$E = 12 / 1.11*10^{-7}=10.75*10^7\ V/m$

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