Answer
a) $V_{ED}$ = $10,000 V$
b) The area is $= 2.26\times10^{-3}m^{2}$
c) The capacitance is $=8.00\times10^{-12}F$
Work Step by Step
a) Since there is a charge on the capacaitors there is electrical energy on the plates and the potential of this energy is given by the equation $V = ED$
$V_{ED}$ = $(4.00\times10^{6} V/m)$ $\times (0.0025 m)$
$V_{ED}$ = $10,000 V$
b) The formula to find area is: $A = Q\divε_{0}E$.
Therefore the area of the plates is $$\frac{80\times10^{-9 }C}{8.854\times10^{-12}C/m^{2}\times4.00\times10^{6} V/m}$$
$$= 2.26\times10^{-3}m^{2}$$
c) The capacitance is given by the formula $C = \frac{ε_{0}A}{d}$.
Therefore the capacitance of the capacitor is
$$\frac{8.854\times10^{-12}F/m \times 2.26\times10^{-3}m^{2}}{00025 m}$$.
$$=8.00\times10^{-12}F$$
