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Chapter 4: Q25E (page 810)

An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 碌C when the potential difference is 200 V. (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.0 xV/m.) (d) When the charge is 0.0180 碌C, what total energy is stored?

Short Answer

Expert verified

(a) The capacitance of an air capacitor of two flat parallel plates apart and having a charge magnitude on each plate of and when the potential difference is 200V is 90pF.

(b) The area of each plate is1.5x10-3m2

(c) If the dielectric breakdowns at3.0x106V/m then the maximum amount of voltage applied without it breaking down is 4.5 kV

(d) The total energy stored when the charge is 0.0180渭Cis1.8x10-6J

Step by step solution

01

Calculating the capacitance of parallel plates

Given,

Charge on each plate Q=0.018渭C, distance between the plates d = 1.5 mm, Potential difference V = 200V and Maximum electric field strengthE6max

Now, the capacitance of a capacitor is given by:

C=QV=0.018x10-6200=90x10-12F=90pF

Therefore, the capacitance of the capacitor is 90pF.

Now capacitance of parallel plate is given by:

C=A蔚鈭�d

A=90脳10-12脳1.5脳10-38.85脳10-12=15.0脳10-3m2

Hence capacitance of the parallel plate is15.0脳10-3m2

02

Calculating the maximum voltage and energy stored

We know that the maximum voltage for a parallel plate is given by:

Vmaxmax=3x106脳1.5x10-3=4.5x103V

Therefore, the maximum voltage is4.5x103V

Now, the energy store in a capacitor is given by:

U=12QV=12脳0.018脳10-6脳200=1.8x10-6J

Therefore, the maximum energy stored is1.8x10-6J

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