A capacitor has parallel plates of area A = 12.6 cm2 separated by d = 3.79 mm .

Question

A capacitor has parallel plates of area A= 12.6 cm2 separated by d= 3.79 mm . The space between the plates is filled with an unknown dielectric with dielectric constant K= 2.40. Find the surface charge density on the plates and the induced surface-charge density on the surface of the dielectric, i when the voltage equals the maximum permissible to avoid dielectric breakdown Vmax= 1.14×105 V . At the threshold of dielectric breakdown, find the surface charge density on the plates and the induced surface-charge density on the surface of the dielectric i. Enter the values for and i separated by commas.

Explanation / Answer

In this case e = epsilon

e = K e0 = 2.40*1.14*10^-11 = 27.32 * 10^-12

Given V max = 1.14*10^5 V

2) E = sigma / e = V /d

sigma = e V /d = 27.32*10^-12 * 1.14*10^5 / 0.00379 = 8.21*10^ -4

3) sigma - sigma = sigma

sigma = e0 V / d = 1.14*10^-11 * 1.14 *10^5 / 0.00379 = 3.42*10^-4

8.21 *10^-4 - 3.42*10^ -4 = 4.79 * 10^-4

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