Respuesta :
(a) Capacitance of the parallel plate capacitor is [tex]3.54\times10^{-9} F[/tex]
(b) Charge on each plate is [tex]3.54\times10^{-6} C[/tex]
(c) Magnitude of the electric field between the plates [tex]2\times 10^{5} N/C[/tex]
It is given that the plates of the parallel-plate capacitor in vacuum are 5.00 mm apart and 2.00 [tex]m^{2}[/tex] in area.
It is also given that a 10.0 kv potential difference is applied across the capacitor.
Capacitance of a parallel plate capacitor when the area of its place and the distance between them is given by,
[tex]C=\frac{A\epsilon_{o}}{d}[/tex]
[tex]C=\frac{2\times\8.85\times10^{-12} }{5\times10^{-3} }[/tex]
[tex]C=3.54\times 10^{-9} F[/tex]
When the capacitance and potential difference applied across it is known, charge on each plate of a parallel plate capacitor is given by,
[tex]q=CV[/tex]
[tex]q=3.54\times10^{-9} \times10000[/tex]
[tex]q=3.54\times10^{-6} C[/tex]
Magnitude of electric field between the plates is given by the expression,
[tex]E=\frac{\sigma}{\epsilon_{o} }=\frac{C}{A\epsilon_{o}}=\frac{CV}{\epsilon_{o}}[/tex]
[tex]E=\frac{3.54\times10^{-9}\times10000 }{8.85\times10^{-12} }[/tex]
[tex]E=2\times10^{5} N/C[/tex]
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