The plates of a parallel-plate capacitor are made of circular discs of radii 5.0 cm each. If the separation between the plates is 1.0 mm, what is the capacitance ?

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Solution

Area of the plate

$A = π r^{2} = π \times (5 \times 10^{- 2})^{2}$

$d = 1.0 \times 10^{- 3} m$

$C = \frac{\in_{0} A}{d}$

$= \frac{8.8 \times 10^{- 12} \times 3.14 \times 25 \times 10^{- 4}}{10^{- 3}}$

$= 695 \times 10^{- 5} μ F$

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The plates of a parallel-plate capacitor are made of circular discs of radii 5.0 cm each. If the separation between the plates is 1.0 mm, what is the capacitance ?

Area of the plate A=πr2=π×(5×10−2)2 d=1.0×10−3 m C=∈0 Ad =8.8×10−12×3.14×25×10−410−3 =695×10−5 μF

Area of the plate

$A = π r^{2} = π \times (5 \times 10^{- 2})^{2}$

$d = 1.0 \times 10^{- 3} m$

$C = \frac{\in_{0} A}{d}$

$= \frac{8.8 \times 10^{- 12} \times 3.14 \times 25 \times 10^{- 4}}{10^{- 3}}$

$= 695 \times 10^{- 5} μ F$