# Capacitance changes by a factor of 3 or 9?

This is an example problem with solution in a book I am reading on electrostatics.

"We have two square parallel plate capacitors with length $a$ and separation $d$ with $d<<a$. If we triple all dimensions of the capacitor, by what factor does the capacitance change?"

$C=\frac{\epsilon_0A}{d}$ and $A$ is the area. The book says the capacitance would change by a factor of $3$, but doesn't it change by a factor of $9$?

$C=\frac{\epsilon_0(a)^2}{d}\longrightarrow\frac{\epsilon_0(3a)^2}{d}= 9\frac{\epsilon_0(a^2)}{d}$

• Hint: What's that $d$ in the denominator? – ACuriousMind Sep 8 '14 at 16:53
• @ACuriousMind: $d$ is the distance between them...? OK, tripling all dimensions also implies the thickness? So the distance between the plates would decrease by a factor of 3? – curiousGeorge119 Sep 8 '14 at 16:56
• Why should the distance decrease if you triple ALL coordinates? And how would that get you a factor of 9? You are pretty close, you just have to plug in the right numbers! – CuriousOne Sep 8 '14 at 16:58
• Tripling all dimensions.. does $d$ become $3d$? If so, then $C=\frac{\epsilon_0(3a)^2}{3d}=\frac{9\epsilon_0(a)^2}{3d}$ for a factor of $3$? – curiousGeorge119 Sep 8 '14 at 17:04