Two parallel plates, each of area 4.00 cm2, are separated by 5.00 mm with purifi

Question

Two parallel plates, each of area 4.00 cm2, are separated by 5.00 mm with purified nonconducting water between them. A voltage of 6.00 V is applied between the plates. Calculate the following.
(a) the magnitude of the electric field between the plates
1.2e3 N/C
(b) the charge stored on each plate
0.340 nC
(c) the charge stored on each plate if the water is removed and replaced with air
nC

I keep getting part C wrong and I don't understand why. I feel that I am doing it correct. Please Help. Thank you

Explanation / Answer

Given Area of the plates is , A = 4.00 cm^2 Distance between the plates is , d = 5.00 mm Voltage applied between the plates is , V = 6.0 V Dielectric constant of water is , k = 80.10 a) The magnitude of theelectric field between the plates is      E = V /d         = 6.0 V / (5.0 *10^-3 m)      E = 1200 V/m ---------------------------------------------------------------------------------- b)The charge stored on each plate is       Q = Cdi V ---- 1            [Cdi = capacitor with dielectric] where as Cdi = k o A /d                  = 80.10 * (8.85 *10^-12 F/m) * (4.0 *10^-4 m^2) /(5.0 *10^-3 m)                Cdi = 5.67 *10^-11 F Taking equation 1 Q = (5.67 *10^-11 F) * 6.0 V Q = 0.34 nC ------------------------------------------------------------------------------------ c) The charge on the plate without dielectric is       Q' = C' V    where C' = Cdi / k = (5.67 *10^-11 F) / 80.10               C' = 7.07 *10^-13 F      Now Q' = (7.07 *10^-13 F) * 6.0 V                    = 4.2 pF

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