A parallel-plate capacitor has capacitance C 0 = 8.30 pF when there is air betwe

Question

A parallel-plate capacitor has capacitanceC0 = 8.30 pF when there is air between the plates. The separation between the plates is 1.30 mm .

What is the maximum magnitude of charge 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 ?

A dielectric with K = 2.80 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 ?

Explanation / Answer

Given: capacitance (C) = 8.3 pF; seperation between plates (d) = 1.3 mm; maximum allowed electric field between plates (Emax) = 3*104 V/m

For the given Emax the potential difference between the plates would be (V) = Emax*d = 3*104*1.3*10-3 = 39V

Hence the charge on the capacitor (Q) = C*V = 39*8.3 pC = 323.7 pC

Now when dielectric with K = 2.8 is inserted then the electric field field in between the plates gets decreased by a factor of K, so the actual eletric field corresponding to maximum electric field of Emax would be K*Emax

So the volate diffenerence would become K*V and thus the charge on the capatitor becomes

Qnew = K*Q = 2.8*323.7 = 906.36 pC

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects