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What is the magnitude of an electrostatic force experienced by one elementary charge at a point in an electric field where the field intensity is 3.0 x 10^3 N/C?

a) 1.0 x 10^3 N
b) 1.6 x 10^-19 N
c) 3.0 x 10^3 N
d) 4.8 x 10^-16 N

Respuesta :

skyluke89

Answer:

d) 4.8 x 10^-16 N

Explanation:

The electrostatic force experienced by a charge is given by

[tex]F=qE[/tex]

where

q is the charge

E is the magnitude of the electric field

In this problem, we have

[tex]q=e=1.6\cdot 10^{-19} C[/tex] (elementary charge)

[tex]E=3.0\cdot 10^3 N/C[/tex] (electric field)

Substituting into the equation, we find the force

[tex]F=(1.6\cdot 10^{-19}C)(3.0\cdot 10^3 N/C)=4.8\cdot 10^{-16} N[/tex]

adeoladokun31

4.8 x 10^-16 N is the magnitude of the Electrostatic force.

The electrostatic force of a charge is given by

F=qE

where

q is the charge

E is the magnitude of the electric field

F is Force

F=1.6 ×10^3

Q= 4.8×10^-16

Therefore,

E= q×F

E= 1.6 ×10^3 ×4.8×10^-16N/C

E= 4.8 x 10^-16 N.

Electrostatic Force is the electric charge around object.

Therefore, The magnitude of the electrostatic Force is 4.8 x 10^-16 N.

Learn more about Electrostatic force from the link below.

https://brainly.com/question/17692887

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