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A positive charge of 4.0 x (10^-5) C experiences a force
of 0.36 N when located at a certain point. What is the
electric field intensity at that point? (Show your work:
Equation, steps and units)

Respuesta :

quasarJose

Answer:

9000N/C

Explanation:

Given parameters:

Charge = 4 x 10⁻⁵C

Magnitude of force  = 0.36N

Unknown:

Electric field intensity = ?

Solution:

This is the force that would be experienced by a unit test charge at a point.

 Electric field intensity  = [tex]\frac{Force}{Charge}[/tex]  

 Insert parameters and solve;

    Electric field intensity  = [tex]\frac{0.36}{4 x 10^{-5} }[/tex]  

    Electric field intensity  = 9000N/C

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