Respuesta :
The magnitude of the displacement current between the plates is [tex]2.1*10^{-8} A[/tex]
Given,
A=4.3*[tex]10^{-2} m^{2}[/tex]
[tex]E=(4.0*10^{5})-(6.0*10^{4}t)\\[/tex]
[tex]i_{d} =ϵ_{0} *\frac{dϕ_{E} }{dt } =ϵ_{0}A\frac{dE}{dt}[/tex]
[tex]i_{d}[/tex]=[tex]ϵ_{0}[/tex]*A*[tex]\frac{d}{dt}[/tex][tex](4.0*10^{5})-(6.0*10^{4}t)[/tex]=[tex]-ϵ_{0} *A*6.0*10^{4}[/tex]
= -[tex](8.85*10^{-12})(4.0×10*^{-2})(6.0×10^{4})[/tex]=[tex]-2.1*10^{-8} A[/tex]
Current
An electrical charge carrier flow known as current often involves electrons or atoms lacking in electrons. The capital letter I is frequently used as a symbol for current. Amperes are the common unit and are denoted by the letter A. A coulomb of electrical charge moves past a certain place in one second as one ampere of current does. Franklin current or conventional current are terms used by physicists to describe how current flows from relatively positive to comparatively negative sites. Negatively charged electrons are the most prevalent charge carriers. They move in a somewhat good direction from relatively negative points.
With e in volts per meter and t in seconds. at t = 0, the field is upward. the plate area is 4. 3 × 10-2 m2. for t > 0, what is the magnitude of the displacement current between the plates?
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