Answer
$\sigma = 4.60\times 10^{-6}~C/m^2$
Work Step by Step
We can find an expression for the capacitance:
$C = \frac{\kappa ~\epsilon_0~A}{d}$
We can find an expression for the charge on the higher-potential plate:
$q = C~V$
$q = (\frac{\kappa~\epsilon_0~A}{d})~(V)$
$q = \frac{\kappa~\epsilon_0~A~V}{d}$
We can find the charge density $\sigma$:
$\sigma = \frac{q}{A}$
$\sigma = \frac{\frac{\kappa~\epsilon_0~A~V}{d}}{A}$
$\sigma = \frac{\kappa~\epsilon_0~V}{d}$
$\sigma = \frac{(2.60)(8.854\times 10^{-12}~C/V~m)(600~V)}{3.00\times 10^{-3}~m}$
$\sigma = 4.60\times 10^{-6}~C/m^2$
