A parallel-plate capacitor has capacitance C0 = 3.00 pF when there is air betwee

Question

A parallel-plate capacitor has capacitance C0 = 3.00 pF when there is air between the plates. The separation between the plates is 1.40 mm.

(a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 104 V/m? ___C

(b) A dielectric with K = 2.60 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 104 V/m? ___C

Explanation / Answer

Here the electric field inside a parallel plate capacitor is:

E = V/d

Also:

Q = C*V

(a) V = E*d = 3.00*10^4 * 0.0014 = 42 V

Q = C*V = 3.0*10^-12 * 42 = 126*10^-12 coul. = 126 pC

(b) The dielectric will multiply the capacitance by the relative dielectric constant.

C = *C = 2.6*3.0 pF = 7.8 pF

Since the electric field hasn't changed, the voltage also has not changed, so the charge is:

Q = C*V = 7.8*10^-12 * 42 = 327.6 *10^-12 coul = 327.6 pC

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