A parallel-plate capacitor has a capacitance of 740 pF, a plate area of 610 cm 2
ID: 2280487 • Letter: A
Question
A parallel-plate capacitor has a capacitance of 740 pF, a plate area of 610 cm2, and a mica dielectric (? = 5.4) completely filling the space between the plates. At 57 V potential difference, calculate (a) the electric field magnitude E (in N/C) in the mica, (b) the magnitude of the free charge on the plates (in C), and (c) the magnitude of the induced surface charge (in C) on the mica.
(a)
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Units
(a)
Number
Units
N/C or V/m V kV mV V/s V/m A parallel-plate capacitor has a capacitance of 740 pF, a plate area of 610 cm2, and a mica dielectric (? = 5.4) completely filling the space between the plates. At 57 V potential difference, calculate (a) the electric field magnitude E (in N/C) in the mica, (b) the magnitude of the free charge on the plates (in C), and (c) the magnitude of the induced surface charge (in C) on the mica.Explanation / Answer
A=610*10^-4 m^2
a)
C=KeoA/d
=>d =KeoA/C =5.4*(8.85*10^-12)*(610*10^-4)/(740*10^-12)
d=3.94*10^-3 m
E=V/d =57/(3.94*10^-3)
E=1.446*10^4 N/C
b)
Qo=CV =740*10^-12*57
Qo=4.218*10^-8 C or 42.18 nC
c)
the magnitude of induced charge
Q =Qo-eoAE =4.218*10^-8-(8.85*10^-12)*(610*10^-4)*(1.446*10^4)
Q=3.44*10^-8 C or 34.4 nC