A 200 pF parallel-plate capacitor has ±35 nC charges on its plates. The plates are separated by 0.3 mm. Find:
a) the area of each plate;
cm2
b) the potential difference between the plates;
V
c) the electric field between the plates;
N/C
Here ,
d = 0.3 mm
C = 200 pF
a) let the area of plates is A
C = A * epsilon/d
200 *10^-12 = 8.854 *10^-12 * A/(.3 *10^-3)
solving for A
A = 6.77 *10^-3 m^2
A = 67.7 cm^2
area of plate is 67.7 cm^2
b)
as Q = C * V
35 *10^-9 = 200 *10^-12 * V
V = 175 V
the potential difference between the plates is 175 V
c)
electric field = potential difference/d
electric field = 175/(.3 *10^-3)
electric field = 5.83 *10^5 V
A 200 pF parallel-plate capacitor has ±35 nC charges on its plates. The plates are separated...
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