Two parallel plates, each of area 3.00cm 2 , are separated by 5.00 mmwith purifi

Question

Two parallel plates, each of area 3.00cm2, are separated by 5.00 mmwith purified nonconducting water between them. A voltage of10.00 V is applied between the plates.Calculate the following. (a) the magnitude of the electric field betweenthe plates
N/C
(b) the charge stored on each plate
nC
(c) the charge stored on each plate if the water is removed andreplaced with air
nC (a) the magnitude of the electric field betweenthe plates
N/C
(b) the charge stored on each plate
nC
(c) the charge stored on each plate if the water is removed andreplaced with air
nC

Explanation / Answer

This is done similarily to your other question: Because we are holding voltage constant (vs. charge) the electricfield is independant of the dielectric, i.e. E=E0 (a) E=V/d E=10V/.005m=2000V/m=2000N/C....N/C and V/m are actually the sameunit :) b)For this we have to take the dielectric into account My book of dielectrics lists them by their k value; which isdefined as /0 , so thats what Ill solve thisproblem with. Because I'm using the ratio, its more convientto do part c first: C=Q0/V=0(A/d) Q0/10=8.854x10-12(.0003/.005) Q0=.005nC ( this is the answer to part c) Q= kQ0 (k for water at room temperature=80.4) Q=80.4(.005)=.402nC

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