Question

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.

Answer 1

The magnitude of the electric field between two charged parallel plates remains constant when the plate separation is reduced from d to d/2, provided the charge density on the plates stays the same. Thus, the magnitude of the electric field halfway between the plates would still be E.

Step-by-step explanation:

The student asked what the magnitude of the electric field would be if the separation between two large, flat, parallel, charged plates is reduced from d to d/2, assuming the charge on the plates remains constant.

According to Essential Knowledge 2.C.5, the electric field (E) between two oppositely charged parallel plates with uniformly distributed electric charge is constant in magnitude and direction at points far from the edges of the plates.

The magnitude of the electric field E between parallel plates is given by the equation E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. Since the surface charge density (σ = Q/A, charge per area) does not change when the separation between the plates is altered and the permittivity of free space (ε₀) is a constant, the magnitude of the electric field between the plates remains the same even if the separation is reduced to d/2.

Therefore, the magnitude of the electric field halfway between the plates when the separation is reduced to d/2 would still be E.

Answer 2

E the electric field remains unchanged.

Step-by-step 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.