91Ó°ÊÓ

$C=\frac{Q}{V_{ab}}={\mathrm{Î\mu }}_0\frac{A}{d}$

C =The Capacitance, $V_{ab}$=The potential difference between the plates, Q= Magnitude of charge on each plate, d=Distance between plates, localid="1668318045119" ${\mathrm{Î\mu }}_{\mathrm{ο}}=8.85×{10}^{-12}F/m$electricconstant, A= Area of each plate.

Definition of capacitance: The capacitance C of a capacitor is the ratio of the magnitude of the charge Q on either conductor to the magnitude of the potential difference between the conductors. The capacitance depends only on the geometry of the capacitor. The greater the capacitance C of a capacitor, the greater the magnitude Q of charge on either conductor for a given potential difference V and hence the greater the amount of stored energy.

Apply: we have a bunch of variables in the above equations. So, in most problems, we are asked to get one of them by the above equation either directly by substitution or by solving the equation indirectly for this variable.

Dielectric constant:

$K=\frac{C}{C_0}$

K= Dielectric constant, C = Final capacitance, $C_0$= original capacitance

$\begin{array}{c}{C}_0=\frac{Q}{V}=\frac{{\mathrm{Î\mu }}_{0}A}{d}\\ K=\frac{C}{C_0},\\ C=K{C}_0=K\frac{{\mathrm{Î\mu }}_{0}A}{d}\end{array}$

$A=1c{m}^2={10}^4{m}^2→$as we are told to get the to get the capacitance per square centimeter

$\mathrm{c}=\left(0.01180\mathrm{x}{10}^{-4}\right)\mathrm{F}=1.18\mathrm{x}{10}^{-6}\mathrm{F}$

$\begin{array}{r}V=KEd\\ E=\frac{V}{Kd}=\frac{85×{10}^{−3}\mathrm{V}}{10×7.5×{10}^{−9}\mathrm{m}}=1.133×{10}^6\mathrm{N}/\mathrm{m}\end{array}$