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

Tiefdruckot · Accepted Answer

Step 1GivenWe are given the electric field between the two plates&amp;nbsp;E0=3.20×105Vm&amp;nbsp;the space between the two electrodes is vacuum. And the electric field with a dielectric material is&amp;nbsp;E=2.50×105&amp;nbsp;V/mRequired(a) We are asked to find the charge density&amp;nbsp;σi&amp;nbsp;on each surface of the dielectric.(b) We want to determine the dielectric constant&amp;nbsp;KSolution(a) When a dielectric material is inserted between the plates while the charge is kept constant with a surface charge density&amp;nbsp;σ, an induced charge of the opposite sign appears on each surface of the dielectric with a charge density&amp;nbsp;σi. The charge density&amp;nbsp;σi&amp;nbsp;on the dielectric material is related to the charge density&amp;nbsp;σ&amp;nbsp;on the electrodes by equation 24.16 in the fromσi=σ(1−1K)The plates&#039; potential difference decreases by a factor K. Thus, the electric field between the plates decreases by the same factor and the dielectric constant is related to the electric field&amp;nbsp;E0&amp;nbsp;of the vacuum and the electric field of the dielectric&amp;nbsp;E&amp;nbsp;by equation 24.14K=E0E=3.20×105Vm2.50×105Vm=1.28Where the charge on the plates is constant and K is a unitless number.Step 2Also, the charge density&amp;nbsp;σ&amp;nbsp;before the dielectric material is related to the applied electric field between two parallel plates byσ=ϵ0E0=(8.854×10−12C2N⋅m2)(3.20×105Vm)=2.833×10−6Cm2Now we can plug our values for&amp;nbsp;σ&amp;nbsp;and&amp;nbsp;K&amp;nbsp;into equation (1) to get the surface charge density on the dielectric&amp;nbsp;σiσi=σ(1−1K)=(2.833×106Cm2)(1−11.28)=0.620×10−6Cm2=0.620μCm2

Jimmy Macias · Accepted Answer

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

Don Sumner · Accepted Answer

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=8.854⋅10−12⋅3.2⋅105=2.833⋅10−6c/m2 σi=σ(1−1K)⇒(induced cgarge density on surface of dielctric) =(2.883⋅10−6)c/m2)(1−11.28)=6.2⋅10−7c/m2 Result a)6.2⋅10−7c/m2 b)K=1.28

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

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