What (in N/C) is the electric field 3.00 m from the center of the terminal of a Van de Graaff with an 8.00 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal?

To solve this problem we will apply the concept related to the electric field defined from the laws of Coulomb. For this purpose we will remember that the electric field is equivalent to the product of the Coulomb constant due to the change of the charge over the squared distance, mathematically this is

Here,

k = Coulomb's constant

r = Distance from center of terminal to point where electric field is to found

q = Excess charge placed on the center of terminal of Van de Graff's generator

Replacing we have that,

Therefore the electric field is

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form