Answer:
a) r=4.24cm d=1 cm
b) [tex]Q=5x10^{-10} C[/tex]
Explanation:
The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field
[tex]V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}[/tex]
[tex]V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m[/tex]
The distance must be the separation the r distance can be find also using
[tex]C=\frac{Q}{V_{ab}}[/tex]
But now don't know the charge these plates can hold yet so
a).
d=0.01m
[tex]C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}[/tex]
[tex]A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}[/tex]
[tex]A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } } \\r=42.55x^{-3}m[/tex]
b).
[tex]C=\frac{Q}{V_{ab}}[/tex]
[tex]Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C[/tex]
