Question

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?

Answer 1

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

$E = \frac{kq}{r^2}$

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,

$E = \frac{(9*10^{9})(8*10^{-3})}{3^2}$

$E = 8*10^6N/C$

Therefore the electric field is $8*10^6N/C$