Question

A 3.3-cm-diameter parallel-plate capacitor has a 2.3mm spacing. The electric field strength inside the capacitor is 1.4

Answer 1

potential differebce=V = 320 V.

C = 8.854 x 10^-12 x A/d
A = pi(3^2)/4 = 9pi/4 cm^2 = 9pi/4 x 10^-4 m^2
d = 2.3 mm = 2.3 x 10^-3 m
C = 8.854 x 10^-12)(9pi/4) x 10^-4/2.3 x 10^-3 = 2.72 x 10^-18 farads
Q + CV = 2.72 x 10^-18 x 320= 8.70x 10^-16coulombs on each plate

Answer 2

$C = \epsilon A / d$

so

V = E * d = (2.3 x 10^-3 x 1.4 x 10^5 )= 1.71 x 10^2 = 322 volts.

C = 8.854 x 10^-12 x A/d
A = pi(r^2) = pi (3.3/2)^2 cm^2 = 8.5529 x 10^-4 m^2
d = 2.3 mm = 2.3 x 10^-3 m
C = 8.854 x 10^-12)( 8.5529 x 10^-4 / 2.3 x 10^-3 ) = 3.29 x 10^-12 farads (3.29pF)
Q = CV = 3.29 x 10^-12 x 322 = 1.060 x 10^-9 coulombs on each plate

Answer 3

Answer 4

C = 8.854 x 10^-12 x A/d
A = pi(3^2)/4 = 9pi/4 cm^2 = 9pi/4 x 10^-4 m^2
d = 2.3 mm = 2.3 x 10^-3 m
C = 8.854 x 10^-12)(9pi/4) x 10^-4/2.3 x 10^-3 = 2.72 x 10^-18 farads
Q =CV = 2.72 x 10^-18 x 320= 8.70x 10^-16coulombs on each plate