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alexdok [17]
3 years ago
11

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?
Physics
1 answer:
Snowcat [4.5K]3 years ago
6 0

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

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Answer: D

Explanation:

Atomic weight is measured by adding the number of protons and neutrons in an atom's nucleus. Argon's atomic number is 18 while potassium's is 19. This means that Argon will always have 18 protons while potassium will always have 19 protons.

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