A parallel-plate capacitor has plates of area3.56×104 m^2. What plate separation is required if the capacitance is to be1590pF? Assume space between the plates is filled withair. -What plate separation is required if the capacitance isto be 1590pF? Assume space between the plates is filled withpaper. A parallel-plate capacitor has plates of area3.56×104 m^2. What plate separation is required if the capacitance is to be1590pF? Assume space between the plates is filled withair. -What plate separation is required if the capacitance isto be 1590pF? Assume space between the plates is filled withpaper.
Explanation / Answer
A parallel plate capacitor has Area A =3.56*10-4 m2 Capacitance of a capacitor is C = 1590 pF a ) space between the plates is filled withair the dielectric constant is k = 1.00059 Capacitance of a capacitor is C = k 0 A /d d = k 0 A / C = 1.00059 * 8.85*10-12 * 3.56*10-4m2 / 1590 * 10 -12 F = 1.9*10-6m b ) space between the plates is filledwith paper the dielectric constant is k = 3.7 Capacitance of a capacitor is C = k 0 A /d d = k 0 A / C = 3.7 * 8.85*10-12 * 3.56*10-4m2 / 1590 * 10 -12 F = 7.3*10-6m Capacitance of a capacitor is C = k 0 A /d d = k 0 A / C = 3.7 * 8.85*10-12 * 3.56*10-4m2 / 1590 * 10 -12 F = 7.3*10-6m