A parallel-plate capacitor has capacitance C 0 = 8.50 pF when there is air betwe
ID: 1514427 • Letter: A
Question
A parallel-plate capacitor has capacitanceC0 = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 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 3.00×104 V/m ?
b.) A dielectric with K = 2.50 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
Part A,
C=K*Epsilon_0*(A/d).
8.5e-12=8.85e-12*(A/.001)
A= 9.60*10^-4 m^3
Q=Epsilon*E*A.
Q = 8.85*10^-12*3*10^4*9.6*10^-4
Q = 2.55*10^-10 C
Part B
C=K*Epsilon_0*(A/d).
8.5e-12=2.5*8.85e-12*(A/.001)
A= 3.840*10^-4 m^3
Q=Epsilon*E*A.
Q = 8.85*10^-12*3*10^4*3.84*10^-4
Q = 1.02*10^-10 C