Question

What is the electric field 5.00 m from the center of the terminal of a Van de Graaff with a 3.00 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal?

A) (1.08 X 10⁵ N/C) outward
B) (6.67 X 10⁴ N/C) inward
C) (1.50 X 10⁵ N/C) outward
D) (4.80 X 10⁴ N/C) inward

Answer 1

The electric field 5.00 m from the center of the terminal of a Van de Graaff with a 3.00 mC charge is approximately 1.08 x 10^5 N/C outward.

Step-by-step explanation:

The electric field 5.00 m from the center of the terminal of a Van de Graaff with a 3.00 mC charge, which is equivalent to that of a point charge at the center of the terminal, can be calculated using the formula:

E = kQ/r²

Where E is the electric field, k is the electrostatic constant (9.00 x 10^9 Nm²/C²), Q is the charge, and r is the distance from the center of the terminal. Substituting the given values, we have:

E = (9.00 x 10^9 Nm²/C²)(3.00 x 10^-3 C)/(5.00 m)²

Simplifying the equation, the electric field is approximately 1.08 x 10^5 N/C outward (A).