If the frequency of the voltage signal impressed on the parallel plate capacitor is doubled from f to 2f, the amplitude of the magnetic field bo at r = a at the mid-plane of the gap between the plates of this capacitor will: increase by a factor of two. remain the same. decrease by a factor of four. decrease by a factor of two. increase by a factor of four.

Increase by a factor of four

Explanation:

Given data

capacitor =  f to 2f

r = a

to find out

capacitor will be

solution

first we calculate the current

displacement current will be = ɛ × electric flux  

displacement current will be  = ɛ × electric flux

electric flex = A × electric field

electric field = voltage / d

electric field = V(o) sin ωt / d

so electric flex

electric flex = A ×  V(o) sin ωt / d

and displacement current will be

displacement current = ɛ × A ×  V(o) sin ωt / d

so the magnetic filed will be here

magnetic filed = μ × current + μ × ɛ d∅/dt

so we can say that frequency will be double when magnetic field increase

so Increase by a factor of four

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