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

Question

Two parallel plates, each of area 3.00cm2, are separated by 3.00 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
1 N/C
(b) the charge stored on each plate
2 nC
(c) the charge stored on each plate if the water is removed andreplaced with air
3 nC (a) the magnitude of the electric field betweenthe plates
1 N/C
(b) the charge stored on each plate
2 nC
(c) the charge stored on each plate if the water is removed andreplaced with air
3 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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