Answer
$1.05\times 10^4\ V/m$
Work Step by Step
It is given that:
Capacitance of capacitor $C =100\ pF = 100 \times 10^{-12}\ C$
Area of plate $A = 100\ cm^2 = 100\times 10^{-4}\ m^2$
Dielectric constant $k=5.4$
Potential difference across capacitor $\Delta V =50\ V$
Formula for the Electric field between two plates of a capacitor separated by distance d is:
$E = \frac{\Delta V}{d}$
Formula for the capacitance having area A and separation d is:
$C = \frac{\epsilon_oA}{d}$
This can be rearranged as:
$d = \frac{\epsilon_oA}{C}$
We substitute expression for $d$ in Electric field E and solve;
$E =\frac{VC}{k\epsilon_oA}$
$E=\frac{(50V)( 100\times 10^{-12} F)}{(5.4)(8.85\times 10^{-12}C^2/Nm^2)(100\times 10^{-4}m^2)}$
$E =1.05\times 10^4\ V/m$
