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.

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.

the answer is b ) square root of 5

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2