Two parallel plates have equal and opposite charges. When the space between the

Donald Johnson 2021-12-17 Answered

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E=3.20×105V/m. When the space is filled with dielectric, the electric field is E=2.50×105V/m. (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

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Expert Answer

Tiefdruckot
Answered 2021-12-18 Author has 46 answers

Step 1
Given
We are given the electric field between the two plates E0=3.20×105Vm the spase between the two electrodes is vacuum. And the electric field with a dielectric material is E=2.50×105 V/m
Reqired
(a) We are asked to finde the charge density σi on each surface of the dielectric.
(b) We want to determine the dielectric constant K
Solution
(a) When a dielectric material is inserted between the plates while the charge is kept constant with a surface charge density σ, an induced charge of the opposite sign appears on each surface of the dielectric with a charge density σi. The charge density σi on the dielectric material is related to the charge density σ on the electrodes by equatoin 24.16 in the from
σi=σ(11K)
The potential difference between the plates decreases by a factor K. Therefore, the electric field between the plates decreases by the same factor and the dielectric constant is related to the electric field E0 of the vacuum and the electric field of the dielectric E by equation 24.14
K=E0E
=3.20×105Vm2.50×105Vm
=1.28
Where K is a unitless and the cahrge on the plates is constant
Step 2
Also, the charge density σ before the dielectric material is related to the applied electric field between two parallel plates by
σ=ϵ0E0
=(8.854×1012C2Nm2)(3.20×105Vm)
=2.833×106Cm2
Now we can plud our values for σ and K into equation (1) to get the surface charge density on the dielectric σi
σi=σ(11K)
=(2.833×106Cm2)(111.28)
=0.620×106Cm2
=0.620μCm2

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Jimmy Macias
Answered 2021-12-19 Author has 30 answers
Step 3
(b) The dielectric constant was calculated in part (a) by knowing the electric field after the dielectric and the electric field before the dielectric and equals
K=1.28
The dielectri constant decreases the potential difference and the electric field, while the capacitance increases.
Result
(a)The surface charge density on the dielectric is σ=0.620μCm2.
(b)K=1.28

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Don Sumner
Answered 2021-12-27 Author has 35 answers

Step 1
C=QVab=ϵ0Ad
Vab=Ed
E=Qϵ0A
K=EE0
(K is the dielectric constant, E is the electric field)
Step 2
b)K=EE0=3.22.5=1.28,
a)σ=ϵ0E0=8.85410123.2105=2.833106c/m2
σi=σ(11K)(induced cgarge density on surface of dielctric)
=(2.883106)c/m2)(111.28)=6.2107c/m2
Result
a)6.2107c/m2
b)K=1.28

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