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Capacitance (C) for a parallel plate capacitor can be calculated by multiplying Epsilon naught (Permittivity of free space by an electric field) with the Area (A) of the plates divided by the Distance between the plates (d). C=e0 * A / d. If the pates of the capacitor are square, how does doubling the side of the plates affect the Capacitance?
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let side = a

so Cold = eo (a^2) / d

when doubling the side it's 2a hence A(new) = 4 a^2

so C new = 4 eo A / d

= 4 * C old

so it's 4 times it's old value
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