Question

The Plates of a parallel plate capacitor are seperated by 1.0mm. (a)What is the area of the plate if the capacitance is 2.5(micro Farad) (b) if the distance of seperation is doubled and the plate size is halved, by what factor does the capacitance change? please explain your answer, thank you

Answer 1

a)we know

C = A/d a) we know 2.5 times 10^-6 = 8.85 times 10^_12 A/1 times 10^-3 A = 282.48m2 b)since distance is doubled and plate size is halved so area remains same C = 8.85 times 10^-12 times 282.48/2 times 10^-3 C = 1.25 times 10^-6 F so, capacitance change by a factor of 2.

$2.5\times10^{-6}=\dfrac{8.85\times10^{-12}A}{1\times10^{-3}}$

A=282.48 m²

b)since distance is doubled and plate size is halved so area remains same

$C=\dfrac{8.85\times10^{-12}\times282.48}{2\times10^{-3}}$

$C={1.25\times10^{-6}F$

so, capacitance change by a factor of 2.