Question

What is the electric field 3.3 m from the center of the terminal of a Van de Graaff with a 7.20 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal?

Answer 1

Answer:

Electric Field is $5.943801*10^6 N/C$

Explanation:

Electric Field:

It originates from positive charge and ends at negative charge.

General Formula for electric Field:

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

where:

k is the Coulomb Constant

q is the charge

r is the distance

Given:

q=7.20 mC

r=3.3 meters

k=$8.99*10^9 N.m^2/C^2$

Find:

Electric Field=?

Solution:

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

$E=\frac{(8.99*10^9)(7.20*10^{-3})}{3.3^2}\\E=5943801.653 N/C\\E=5.943801653*10^6 N/C$

Electric Field is $5.943801*10^6 N/C$