A parallel-plate capacitor has a capacitance of C0 = 5.40 pF when there is air b

Question

A parallel-plate capacitor has a capacitance of C0 = 5.40 pF when there is air between the plates. The separation between the plates is 1.25 mm . (a) 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 3.00×10^4 V/m ? (b) A dielectric with a dielectric constant of 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 3.00×10^4 V/m ? ****Please provide concise equations to solve this

Explanation / Answer

A. Given

E = 3*10^4 V/m

C0 = 5.4 pF

d = 1.25 mm = 1.25*10^-3 m

Electric field = potential/d

E = V/d

Q = CV

from above equations:

E = Q/Cd

Q = E*C*d

Q = 3*10^4*5.4*10^-12*1.25*10^-3

Q = 2.025*10^-10 C = 202.5 pC

B.

C = k*e0*A/d

Now

C1 = k*Co = 2.6*5.4*10^-12 = 14.04 pF

rest procedure is same

Q = E*C1*d

Q = 3*10^4*14.04*10^-12*1.25*10^-3 = 5.265*10^-10

Q = 526.5 pF

Let me know if you have any doubt.

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