A parallel-plate capacitor has a capacitance of 740 pF, a plate area of 610 cm 2

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)

Number

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

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