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Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half way between the two plates the electric field has magnitude E. If the separation of the plates is reduced to d/2 what is the magnitude of the electric field half way between the plates? Assume the charge of plates is constant.

Respuesta :

Answer:

E the electric field remains unchanged.

Explanation:

potential difference between the plates of a capacitor

V = Q / C  Where Q is charge on the capacitor and C is capacity of capacitor

Here Q is unchanged.

C the capacity has value equal to ε A / d

Here d is distance between plates. when  it is halved, capacitance C becomes double.

V = Q /C , When C becomes double V becomes half

E = V/d , E is electric field between plates having separation of d.

When V becomes half and d also becomes half , there is no change in the value of E.  

Hence E the electric field remains unchanged.

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