Respuesta :
Given data:
* The area of the parallel plate capacitor is,
[tex]\begin{gathered} A=2cm^2 \\ A=2\times10^{-4}m^2^{} \end{gathered}[/tex]* The distance between the plates is,
[tex]\begin{gathered} d=2\text{ mm} \\ d=2\times10^{-3}\text{ m} \end{gathered}[/tex]Solution:
(a). The capacitance of the capacitor in terms of area and distance between the plates is,
[tex]C=\frac{\epsilon_{\circ}A}{d}[/tex][tex]\text{where }\epsilon_{\circ}\text{ is the electrical permittivity of the fr}ee\text{ spaces}[/tex]Substituting the known values,
[tex]\begin{gathered} C=8.85\times10^{-12}\times\frac{2\times10^{-4}}{2\times10^{-3}} \\ C=8.85\times10^{-13}\text{ F} \end{gathered}[/tex]Thus, the value of the capacitanc is 8.85 times 10 power -13 Farad.
(b). The voltage across the battery is,
[tex]V=6\text{ Volts}[/tex]The charge stored in the capacitor in terms of the voltage and the capacitance is,
[tex]\begin{gathered} C=\frac{Q}{V} \\ Q=CV \end{gathered}[/tex]where Q is the charge stored in the capacitor
Substituting the known values,
[tex]\begin{gathered} Q=8.85\times10^{-13}\times6 \\ Q=53.1\times10^{-13}\text{ Coulomb} \end{gathered}[/tex]Thus, the charge stored in the parallel plate capacitor is 53.1 times 10 power -13 coulomb.
